Construct a right triangle whose base is ^@17 \space cm^@ and su | Math Question and Answer | Edugain Italy

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Construct a right triangle whose base is $17 \space cm$ and sum of its hypotenuse and another side is $23 \space cm$ .

Answer:

Step by Step Explanation:

Let

A B C

$ABC$ be the required triangle such that

B C = 17 c m, ∠ B = 90^{\circ}

$BC = 17 \space cm, \angle B = 90 ^ \circ$ and

A C + A B = 23 c m

$AC + AB = 23 \space cm$ . Draw a

B C = 17 c m .

$BC = 17 \space cm.$

At $B$ construct $\angle CBX = 90^\circ.$
From $B$ cut off $BD = 23 \space cm$ .
Join $CD$ and draw perpendicular bisector of $CD$ intersecting $BD$ at $A$ .
Join $AC$ to get the required triangle $ABC$ .

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