| Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
Exercise 5.4
Question 1
In two similar triangles ABC and DEF, AC = 3 cm and DF = 5 cm. Find the ratio of the areas of the two triangles.
Sol :Given: ΔABC ~ ΔDEF and AC = 3 cm and DF = 5 cm
To find: Areas of the two triangles
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AC})^{2}}{(\mathrm{DF})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(3)^{2}}{(5)^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D E F)}=\frac{9}{25}$
Sol :
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AC})^{2}}{(\mathrm{DF})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(3)^{2}}{(5)^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D E F)}=\frac{9}{25}$
Question 2
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
Given: AM = 6cm and DN = 9cm
Here, ΔABC and ΔDEF are similar triangles
We know that, in similar triangles, corresponding angles are in the same ratio.
⇒∠A = ∠D, ∠B = ∠E and ∠C = ∠F ……(i)
In ΔABM and ΔDEN
∠B = ∠E [from (i)]
and ∠M = ∠N [each 90°]
∴ ΔABC ~ ΔDEF [by AA similarity]
So, $\frac{A M}{D N}=\frac{A B}{D E}=\frac{B M}{E N}$ ……(ii)
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AM})^{2}}{(\mathrm{DN})^{2}}$ [from (ii)]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(6)^{2}}{(9)^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{36}{81}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{4}{9}$
Question 3
In the given figure, ΔABC and ΔDEF are similar, BC = 3cm, EF = 4 cm and area of ΔABC = 54 sq cm. Determine the area of ΔDEF.
Sol :
Given: ΔABC ~ ΔEF, BC = 3cm, EF = 4 cm
and area of ΔABC = 54 sq cm
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{BC})^{2}}{(\mathrm{EF})^{2}}$
$\Rightarrow \frac{54}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(3)^{2}}{(4)^{2}}$ [given]
$\Rightarrow \frac{54}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{9}{16}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{DEF})=\frac{54 \times 16}{9}$
⇒ ar (∆DEF) = 96cm2
Question 4
If ΔABC ~ ΔDEF, AB =10 cm, area (ΔABC) = 20 sq. cm, area (ΔDEF) = 45 sq. cm. Determine DE.
Sol :
Given: ΔABC ~ΔDEF, AB = 10cm,
and area of ΔABC = 20 sq cm , area of ΔDEF = 45 sq cm
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{20}{45}=\frac{(10)^{2}}{(\mathrm{DE})^{2}}$ [given]
$\Rightarrow \frac{20}{45}=\frac{100}{(D E)^{2}}$
$\Rightarrow(D E)^{2}=\frac{100 \times 45}{20}$
⇒(DE)2 = 5 × 45
⇒ DE = 15cm
Question 5
In ΔABC ~ ΔADE and DE|| BC. If DE = 3cm, BC = 6 cm and area (ΔADE) =15 sq. cm, find the area of ΔABC.
Sol :
Given: ΔABC ~ ΔADE
DE = 3cm, BC = 6 cm and area (ΔADE) =15 sq. cm
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$
$\Rightarrow \frac{15}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(3)^{2}}{(6)^{2}}$ [given]
$\Rightarrow \frac{15}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{9}{36}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{ABC})=\frac{15 \times 36}{9}$
⇒ ar (∆ABC) = 60cm2
Question 6
In the figure DE || BC. If DE = 4 cm, BC = 8 cm and area (ΔADE) = 25 sq. cm, find the area of ΔABC.
Sol :
Given: DE || BC
DE = 4cm, BC = 8cm and area (ΔADE) =25 sq. cm
In ΔABC andΔADE
∠B = ∠D [∵ DE || BC and AB is transversal,
Corresponding angles are equal]
∠C = ∠E [∵ DE || BC and AC is transversal,
Corresponding angles are equal]
∠BAC =∠DAE [common angle]
∴ ΔABC ~ΔADE [by AAA similarity]
Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$
$\Rightarrow \frac{25}{\operatorname{ar}(\Delta A B C)}=\frac{(4)^{2}}{(8)^{2}}$ [given]
$\Rightarrow \frac{25}{\operatorname{ar}(\Delta A B C)}=\frac{16}{64}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{ABC})=\frac{25 \times 64}{16}$
⇒ ar(ΔABC) = 25×4
⇒ ar (ΔABC) = 100cm2
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$
$\Rightarrow \frac{25}{\operatorname{ar}(\Delta A B C)}=\frac{(4)^{2}}{(8)^{2}}$ [given]
$\Rightarrow \frac{25}{\operatorname{ar}(\Delta A B C)}=\frac{16}{64}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{ABC})=\frac{25 \times 64}{16}$
⇒ ar(ΔABC) = 25×4
⇒ ar (ΔABC) = 100cm2
Question 7
Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. find the ratio of their corresponding heights.
Sol :
Let ΔABC and ΔDEF are two isosceles triangles with AB =AC and DE = DF and ∠A = ∠D
Now, let AM and DN are their respective altitudes or heights.
Let ΔABC and ΔDEF
$\frac{A B}{D E}=\frac{A C}{D F}$
∠A = ∠D [given]
∴ ΔABC ~ΔDEF [by SAS similarity]
We know that, in similar triangles, corresponding angles are in the same ratio.
⇒∠B = ∠E and ∠C = ∠F ……(i)
In ΔABM and ΔDEN
∠B = ∠E [from (i)]
and ∠M = ∠N [each 90°]
∴ ΔABC ~ ΔDEF [by AA similarity]
So, $\frac{A M}{D N}=\frac{A B}{D E}=\frac{B M}{E N}$ ……(ii)
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{16}{25}=\frac{(A M)^{2}}{(D N)^{2}}$ [from (ii)]
$\Rightarrow \frac{(4)^{2}}{(5)^{2}}=\frac{(A M)^{2}}{(D N)^{2}}$
$\Rightarrow \frac{A M}{D N}=\frac{4}{5}$
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{16}{25}=\frac{(A M)^{2}}{(D N)^{2}}$ [from (ii)]
$\Rightarrow \frac{(4)^{2}}{(5)^{2}}=\frac{(A M)^{2}}{(D N)^{2}}$
$\Rightarrow \frac{A M}{D N}=\frac{4}{5}$
Question 8
The areas of two similar triangles are 100 cm2 and 49 cm2, respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.
Sol :
Given: Let ΔABC = 100cm2 and ΔDEF = 49cm2
Let AM = 5cm
Here, ΔABC and ΔDEF are similar triangles
We know that, in similar triangles, corresponding angles are in the same ratio.
⇒∠B = ∠E and ∠C = ∠F …(i)
In ΔABM and ΔDEN
∠B = ∠E [from (i)]
and ∠M = ∠N [each 90°]
∴ ΔABC ~ ΔDEF [by AA similarity]
So, $\frac{A M}{D N}=\frac{A B}{D E}=\frac{B M}{E N}$ ……(ii)
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{100}{49}=\frac{5^{2}}{(\mathrm{DN})^{2}}$ [from (ii)]
$\Rightarrow \frac{100}{49}=\frac{25}{(\mathrm{DN})^{2}}$
$\Rightarrow(\mathrm{DN})^{2}=\frac{25 \times 49}{100}$
$\Rightarrow(D N)^{2}=\frac{49}{4}$
$\Rightarrow \mathrm{DN}=\frac{7}{2}$
⇒DN = 3.5cm
The height of the other altitude is 3.5cm
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{100}{49}=\frac{5^{2}}{(\mathrm{DN})^{2}}$ [from (ii)]
$\Rightarrow \frac{100}{49}=\frac{25}{(\mathrm{DN})^{2}}$
$\Rightarrow(\mathrm{DN})^{2}=\frac{25 \times 49}{100}$
$\Rightarrow(D N)^{2}=\frac{49}{4}$
$\Rightarrow \mathrm{DN}=\frac{7}{2}$
⇒DN = 3.5cm
The height of the other altitude is 3.5cm
Question 9
The areas of two similar triangles are 100 cm2 and 64cm2 respectively. If a median of the smaller triangle is 5.6 cm, find the corresponding median of the other.
Sol :
Let ΔABC and ΔDEF are two similar triangles such that ar (ΔABC) =100cm2 and ar (ΔDEF) = 64cm2
Also, let AM and DN are medians of ΔABC and ΔDEF respectively.
Now in ΔABC and ΔDEF
∠B = ∠E [∵ΔABC ~ ΔDEF]
and $\frac{A B}{D E}=\frac{B M}{E N}$ $\left[\because \frac{A B}{D E}=\frac{B C}{E F} \Rightarrow \frac{A B}{D E}=\frac{2 B M}{2 E N}\right]$
∴ ΔABC ~ ΔDEF [by SAS similarity]
$\Rightarrow \frac{A B}{D E}=\frac{A M}{D N}$ …(i)
∠B = ∠E [∵ΔABC ~ ΔDEF]
and $\frac{A B}{D E}=\frac{B M}{E N}$ $\left[\because \frac{A B}{D E}=\frac{B C}{E F} \Rightarrow \frac{A B}{D E}=\frac{2 B M}{2 E N}\right]$
∴ ΔABC ~ ΔDEF [by SAS similarity]
$\Rightarrow \frac{A B}{D E}=\frac{A M}{D N}$ …(i)
Now, as ΔABC ~ ΔDEF
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{(\mathrm{AM})^{2}}{(\mathrm{DN})^{2}}$ [from (i)]
$\Rightarrow \frac{100}{64}=\frac{(A M)^{2}}{(5.6)^{2}}$
$\Rightarrow(\mathrm{AM})^{2}=\frac{100 \times 5.6 \times 5.6}{64}$
$\Rightarrow(\mathrm{AM})^{2}=\frac{100 \times 56 \times 56}{64 \times 10 \times 10}$
$\Rightarrow(\mathrm{AM})^{2}=7 \times 7$
⇒AM = 7cm
Hence, the length of the other median is 7cm.
Sol :
Given: DE || BC
DE = 5cm, BC = 10cm and area (ΔADE) =20 sq. cm
In ΔABC andΔADE
∠B = ∠D [∵ DE || BC and AB is transversal,
Corresponding angles are equal]
∠C = ∠E [∵ DE || BC and AB is transversal,
Corresponding angles are equal]
∠BAC =∠DAE [common angle]
∴ ΔABC ~ΔADE [by AAA similarity]
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{(\mathrm{AM})^{2}}{(\mathrm{DN})^{2}}$ [from (i)]
$\Rightarrow \frac{100}{64}=\frac{(A M)^{2}}{(5.6)^{2}}$
$\Rightarrow(\mathrm{AM})^{2}=\frac{100 \times 5.6 \times 5.6}{64}$
$\Rightarrow(\mathrm{AM})^{2}=\frac{100 \times 56 \times 56}{64 \times 10 \times 10}$
$\Rightarrow(\mathrm{AM})^{2}=7 \times 7$
⇒AM = 7cm
Hence, the length of the other median is 7cm.
Question 10
In the given figure, DE || BC. If DE = 5 cm, BC =10 cm and ar(ΔADE) = 20 cm2, find the area of ΔABC.
Sol :
Given: DE || BC
DE = 5cm, BC = 10cm and area (ΔADE) =20 sq. cm
In ΔABC andΔADE
∠B = ∠D [∵ DE || BC and AB is transversal,
Corresponding angles are equal]
∠C = ∠E [∵ DE || BC and AB is transversal,
Corresponding angles are equal]
∠BAC =∠DAE [common angle]
∴ ΔABC ~ΔADE [by AAA similarity]
Now, we know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$
$\Rightarrow \frac{20}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(5)^{2}}{(10)^{2}}$ [given]
$\Rightarrow \frac{20}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{25}{100}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{ABC})=\frac{20 \times 100}{25}$
⇒ ar(ΔABC) = 20×4
⇒ ar (ΔABC) = 80cm2
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$
$\Rightarrow \frac{20}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(5)^{2}}{(10)^{2}}$ [given]
$\Rightarrow \frac{20}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{25}{100}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{ABC})=\frac{20 \times 100}{25}$
⇒ ar(ΔABC) = 20×4
⇒ ar (ΔABC) = 80cm2
Question 11
The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the other.
Sol :
Given: Let ΔABC = 81cm2 and ΔDEF = 49cm2
Let AM = 6.3cm
Here, ΔABC and ΔDEF are similar triangles
We know that, in similar triangles, corresponding angles are in the same ratio.
⇒∠B = ∠E and ∠C = ∠F …(i)
In ΔABM and ΔDEN
Here, ΔABC and ΔDEF are similar triangles
We know that, in similar triangles, corresponding angles are in the same ratio.
⇒∠B = ∠E and ∠C = ∠F …(i)
In ΔABM and ΔDEN
∠B = ∠E [from (i)]
and ∠M = ∠N [each 90°]
and ∠M = ∠N [each 90°]
∴ ΔABC ~ ΔDEF [by AA similarity]
So, $\frac{A M}{D N}=\frac{A B}{D E}=\frac{B M}{E N}$ …(ii)
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{81}{49}=\frac{(6.3)^{2}}{(\mathrm{DN})^{2}}$ [from (ii)]
$\Rightarrow \frac{81}{49}=\frac{6.3 \times 6.3}{(\mathrm{DN})^{2}}$
$\Rightarrow(\mathrm{DN})^{2}=\frac{6.3 \times 6.3 \times 49}{81}$
$\Rightarrow(D N)^{2}=\frac{63 \times 63 \times 49}{81 \times 10 \times 10}$
$\Rightarrow(D N)^{2}=\frac{7 \times 7 \times 49}{100}$
⇒DN = 4.9cm
Height of the other altitude is 4.9cm
Sol :
Given: ΔABC ~ ΔDEF and AB =2DE
And area of ΔABC is 56 sq. cm
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{56}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(2 \mathrm{DE})^{2}}{(\mathrm{DE})^{2}}$ [given]
$\Rightarrow \frac{56}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{4(\mathrm{DE})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{DEF})=\frac{56}{4}$
⇒ ar(∆DEF) = 14sq cm
Sol :
It is given that DE || BC and DE : BC = 4 : 5
Let ∆ ADE and ∆ABC
∠ADE = ∠ABC [corresponding angles]
∠AED = ∠ACB [corresponding angles]
∴ ∆ ADE ~ ∆ABC [by AA similarity]
We know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{(\mathrm{BC})^{2}}{(\mathrm{DE})^{2}}$
Subtracting 1 from both the sides, we get
$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D E)}-1=\frac{(5)^{2}}{(4)^{2}}-1$ [given]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})-\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{25-16}{16}$
$\Rightarrow \frac{\operatorname{ar}(\mathrm{CEDB})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{9}{16}$
⇒ $\frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\mathrm{CEDB})}=\frac{16}{9}$
Sol :
Given: AP= 1 cm, 1 BP= 3 cm, AQ = 1.5 cm, CQ = 4.5 cm
Here, $\frac{A P}{P B}=\frac{1}{3}$ and $\frac{A Q}{Q C}==\frac{1.5}{4.5}=\frac{1}{3}$
⇒PQ || BC [by converse of basic proportionality theorem]
In ΔABC andΔAPQ
∠B = ∠P [∵ PQ || BC and AB is transversal,
Corresponding angles are equal]
∠C = ∠Q [∵ PQ || BC and AC is transversal,
Corresponding angles are equal]
∠BAC =∠PAQ [common angle]
∴ ΔABC ~ΔAPQ [by AAA similarity]
Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{AP})^{2}}{(\mathrm{AB})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(1)^{2}}{(1+3)^{2}}$ [given]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{1}{16}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{APQ})=\frac{1}{16} \operatorname{ar}(\Delta \mathrm{ABC})$
Hence Proved
Sol :
Given: AD ⊥ BC
and BC = 13 cm and AC = 5 cm
Let ΔABC and ΔADC
∠A = ∠D [each 90°]
∠C = ∠C [common angle]
∴ ΔABC ~ ΔADC [by AA similarity]
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{(\mathrm{BC})^{2}}{(\mathrm{AC})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D C)}=\frac{(13)^{2}}{(5)^{2}}$
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{81}{49}=\frac{(6.3)^{2}}{(\mathrm{DN})^{2}}$ [from (ii)]
$\Rightarrow \frac{81}{49}=\frac{6.3 \times 6.3}{(\mathrm{DN})^{2}}$
$\Rightarrow(\mathrm{DN})^{2}=\frac{6.3 \times 6.3 \times 49}{81}$
$\Rightarrow(D N)^{2}=\frac{63 \times 63 \times 49}{81 \times 10 \times 10}$
$\Rightarrow(D N)^{2}=\frac{7 \times 7 \times 49}{100}$
⇒DN = 4.9cm
Height of the other altitude is 4.9cm
Question 12
In the given figure, ΔABC ~ ΔDEF. If AB = 2DE and area of ΔABC is 56 sq. cm, find the area of ΔDEF.
Sol :
Given: ΔABC ~ ΔDEF and AB =2DE
And area of ΔABC is 56 sq. cm
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(\mathrm{AB})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \frac{56}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{(2 \mathrm{DE})^{2}}{(\mathrm{DE})^{2}}$ [given]
$\Rightarrow \frac{56}{\operatorname{ar}(\Delta \mathrm{DEF})}=\frac{4(\mathrm{DE})^{2}}{(\mathrm{DE})^{2}}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{DEF})=\frac{56}{4}$
⇒ ar(∆DEF) = 14sq cm
Question 13
In the given figure, DE || BC and DE : BC = 4 : 5. Calculate the ratio of the areas of ΔADE and the trapezium ΔCEDB.
Sol :
It is given that DE || BC and DE : BC = 4 : 5
Let ∆ ADE and ∆ABC
∠ADE = ∠ABC [corresponding angles]
∠AED = ∠ACB [corresponding angles]
∴ ∆ ADE ~ ∆ABC [by AA similarity]
We know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{(\mathrm{BC})^{2}}{(\mathrm{DE})^{2}}$
Subtracting 1 from both the sides, we get
$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D E)}-1=\frac{(5)^{2}}{(4)^{2}}-1$ [given]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})-\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{25-16}{16}$
$\Rightarrow \frac{\operatorname{ar}(\mathrm{CEDB})}{\operatorname{ar}(\Delta \mathrm{ADE})}=\frac{9}{16}$
⇒ $\frac{\operatorname{ar}(\Delta \mathrm{ADE})}{\operatorname{ar}(\mathrm{CEDB})}=\frac{16}{9}$
Question 14
ABC is a triangle, and PQ is a straight line meeting AB in P and AC in Q. If AP= 1 cm, 1 BP = 3 cm, AQ = 1.5 cm, CQ = 4.5 cm. Prove that the area of Δ APQ = 1/16 (area of ΔABC).
Sol :
Given: AP= 1 cm, 1 BP= 3 cm, AQ = 1.5 cm, CQ = 4.5 cm
Here, $\frac{A P}{P B}=\frac{1}{3}$ and $\frac{A Q}{Q C}==\frac{1.5}{4.5}=\frac{1}{3}$
⇒PQ || BC [by converse of basic proportionality theorem]
In ΔABC andΔAPQ
∠B = ∠P [∵ PQ || BC and AB is transversal,
Corresponding angles are equal]
∠C = ∠Q [∵ PQ || BC and AC is transversal,
Corresponding angles are equal]
∠BAC =∠PAQ [common angle]
∴ ΔABC ~ΔAPQ [by AAA similarity]
Now, we know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(\mathrm{AP})^{2}}{(\mathrm{AB})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{(1)^{2}}{(1+3)^{2}}$ [given]
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{APQ})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{1}{16}$
$\Rightarrow \operatorname{ar}(\Delta \mathrm{APQ})=\frac{1}{16} \operatorname{ar}(\Delta \mathrm{ABC})$
Hence Proved
Question 15
ΔABC is right angled at A and AD ⊥ BC. If BC = 13 cm and AC = 5 cm, find the ratio of the areas of ΔABC and ΔADC.
Sol :
Given: AD ⊥ BC
and BC = 13 cm and AC = 5 cm
Let ΔABC and ΔADC
∠A = ∠D [each 90°]
∠C = ∠C [common angle]
∴ ΔABC ~ ΔADC [by AA similarity]
We know that, the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{(\mathrm{BC})^{2}}{(\mathrm{AC})^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta A D C)}=\frac{(13)^{2}}{(5)^{2}}$
$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{ADC})}=\frac{169}{25}$
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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