[4 marks sum]

Question 14 Marks

D and E are points on the sides AB and AC respectively of a Δ ABC such that DE | | BC and divides Δ ABC into two parts, equal in area. Find $\frac{ BD }{ AB }$.

Answer

We have
area $(\triangle ADE) =$ area (trapezium BCED)
$\Rightarrow$ area $(\triangle ADE) +$ area $(\triangle ADE)$
= area (trapezium BCED) + area $(\triangle ADE)$
$\Rightarrow 2$ area $(\triangle ADE) =$ area $(\triangle ABC) ...(i)$
In $\triangle ADE$ and $\triangle ABC,$ we have
$\angle ADE = \angle B, ...[\because DE | | BC]$
$\therefore \angle AED = \angle C $...(corresponding angles)]
and $\angle A = \angle A,$ ...[Common]

∴ Δ ADE ∼ Δ ABC
$\Rightarrow \frac{\operatorname{area}(\Delta ADE )}{\operatorname{area}(\Delta ABC )}=\frac{ AD ^2}{ AB ^2}$
$\Rightarrow \frac{\operatorname{area}(\Delta ADE )}{\text { 2area }(\Delta ADE )}=\frac{ AD ^2}{ AB ^2}$
$\Rightarrow \frac{1}{2}=\left(\frac{ AD }{ AB }\right)^2$
$\Rightarrow \frac{A D}{A B}=\frac{1}{\sqrt{2}}$
$\Rightarrow A B=\sqrt{2} A D$
$\Rightarrow A B=\sqrt{2}(A B-B D)$
$\Rightarrow(\sqrt{2}-1) A B=\sqrt{2} B D$
$\Rightarrow \frac{ BD }{ AB }=\frac{\sqrt{2}-1}{\sqrt{2}}$
$=\left(2-\frac{\sqrt{2}}{2}\right)$

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Question 24 Marks

Equilateral triangles are drawn on the sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

Answer

Given A right angled triangle ABC with right angle at B. Equilateral triangles PAB, QBC and RAC are described on sides AB, BC and CA respectively.
To Prove.
Area (ΔPAB) + Area (ΔQBC) = Area (ΔRAC).
Proof. Since, triangles PAB, QBC and RAC are equilateral. Therefore they are equiangular and hence similar.

$\therefore \frac{\operatorname{area}(\Delta PAB )}{\text { area }(\Delta RAC )}+\frac{\operatorname{area}(\Delta QBC )}{\operatorname{area}(\Delta RAC )}$
$=\frac{ AB ^2}{ AC ^2}+\frac{ BC ^2}{ AC ^2}$
$=\frac{ AB ^2+ BC ^2}{ AC ^2}$
$=\frac{ AC ^2}{ AC ^2}=1$
$\left[\because \triangle ABC\right.$ is a right angled triangle with $\left.\angle B =90^{\circ} \therefore AC ^2= AB^2+ BC ^2\right]$
$\Rightarrow \frac{\operatorname{area}(\Delta PAB )+\operatorname{area}(\Delta QBC )}{\operatorname{area}(\Delta RAC )}=1$
$\Rightarrow$ area (ΔPAB) + area (ΔQBC)
$=$ area (ΔRAC).

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Question 34 Marks

In Δ ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
1) Prove that ΔACD is similar to ΔBCA.
2) Find BC and CD
3) Find area of ΔACD: area of ΔBCA

Answer

∠ABC = ∠DAC = x(say)

AB = 8 cm,
AC = 4 cm,
AD = 5 cm.
(i) In ΔACD and ΔBCA
∠ABC = ∠DAC ...(Given)
∠ACD = ∠BCA ...(Common angles)
⇒ ΔACD ∼ ΔBCA ...(AA criterion for similarity)(i).
Hence ΔACD is similar to ΔBCA.
(ii) In ΔACD and ΔBCA
ΔACD ∼ ΔBCA ...[from (i)]
$\frac{ AC }{ BC }=\frac{ CD }{ CA }=\frac{ AD }{ BA }$
$\Rightarrow \frac{4}{ BC }=\frac{ CD }{4}=\frac{5}{8}$
$\Rightarrow \frac{4}{ BC }=\frac{5}{8}$
$\Rightarrow BC =\frac{8 \times 4}{5}=\frac{32}{5}$
$=6.4 cm .$
$\text { and } \frac{C D}{4}=\frac{5}{8}$
$\Rightarrow C D=\frac{5 \times 4}{8}$
$\Rightarrow C D=2.5 cm .$
(iii) In ΔACD and ΔBCA
ΔACD ∼ ΔBCA ...[from (i)]
$\frac{\text { Area of } \triangle ACD }{\text { Area of } \triangle ABC }=\left(\frac{ AC }{ AB }\right)^2$
$\Rightarrow \frac{\text { Area of } \triangle ACD }{\text { Area of } \triangle ABC }=\frac{5^2}{8^2}=\frac{25}{64}$

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Mathematics [4 marks sum]

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