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Similarity — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsSimilarity3 Marks
Question
$ABC$ is a right angled triangle with $\angle ABC = 90^\circ. D$ is any point on $AB$ and $DE$ is perpendicular to $AC.$ Prove that
Find, area of $\triangle ADE :$ area of quadrilateral $BCED.$
✓

Answer

We need to find the area of $\square ADE$ and quadrilateral $BCED$
Area of $\triangle ADE =\frac{1}{2} \times A E X X D E=\frac{1}{2} \times 4 \times \frac{5}{3}=\frac{10}{3} cm ^3$
Area of $\quad.BCED =$ Area of $ΔABC$ Area of $ΔADE$
$=\frac{1}{2} \times B C \times A B-\frac{10}{3}$
$=\frac{1}{2} \times 5 \times 12-\frac{10}{3}$
$=30-\frac{10}{3}$
$=\frac{80}{3} cm ^2$
Thus ratio of areas of $ADE$ to quadrilateral $BCED =\frac{\frac{10}{3}}{\frac{80}{3}}=\frac{1}{8}$

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