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Quadrilaterals — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsQuadrilaterals4 Marks
Question
In figure, Triangle ABC is a right angled triangle at B. Given that AB = 9cm, AC = 15cm and D, E are the mid-points of the sides AB and AC respectively, calculate
  1. The length of BC
  2. The area of $\triangle\text{ADE}.$
✓

Answer


i. In $\triangle ABC , \angle B =90^{\circ}$,
By using Pythagoras theorem
$A C^2=A B^2+B C^2$
$\Rightarrow 15^2=9^2+BC^2$
$\Rightarrow BC=\sqrt{15^2-9^2}$
$\Rightarrow BC=\sqrt{225-81}$
$\Rightarrow BC=\sqrt{144}=12 cm$
ii. In $\triangle ABC$,
$D$ and $E$ are mid-points of $A B$ and $A C$
$\therefore DE \| BC ,=\frac{1}{2} BC [ By$ mid-point theorem]
$AD = DB =\frac{ AB }{2}=\frac{9}{2}=4.5 cm[\therefore D$ is the mid-point of AB $]$
Area of $\triangle ADE =\frac{1}{2} \times AD \times DE$
$=\frac{1}{2} \times 4.5 \times 6$
$=13.5 cm^2$

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