Construct $△ M A P$ such that $M P = 5 c m, A P = 5 c m a n d ∠ P = 90 °$ . Measure other angles of the of the triangle. What type of triangle is this?

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Solution

STEPS OF CONSTRUCTION:
STEP 1: First we draw a line segment $M P$ of length $5 c m$ .
STEP 2: Now we make a ray $P X$ making an angle of $∠ M P X = 90 °$ .
STEP 3: With $P$ as centre and $5 c m$ as radius draw an arc intersecting $P X$ at $A$ .
STEP 4: Now join $A M$ and we get the required triangle $△ A M P$ .
Now two sides of this triangle are equal it is an isosceles triangle. So the other two angles must be equal.
By angle sum property of triangle,
$∠ M A P + ∠ M P A + ∠ A M P = 180 ° x + 90 ° + x = 180 ° 2 x + 90 ° = 180 ° 2 x = 180 ° - 90 ° 2 x = 90 ° x = \frac{90 °}{2} = 45 °$
$∠ A M P = ∠ P A M = 45 °$

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Construct △MAP such that MP=5cm,AP=5cmand∠P=90°. Measure other angles of the of the triangle. What type of triangle is this?

Construct $△ M A P$ such that $M P = 5 c m, A P = 5 c m a n d ∠ P = 90 °$ . Measure other angles of the of the triangle. What type of triangle is this?