Question

right triangles mnp and qrs are congruent. what is the area of △mnp? 40 m² 60 m² 68 m² 127.5 m²

Explanation:

Step1: Recall congruent triangles property

Congruent triangles have equal corresponding sides and areas. So, we can find a leg of \( \triangle MNP \) using \( \triangle QRS \). In \( \triangle QRS \), one leg is \( 8 \, \text{m} \), and since \( \triangle MNP \cong \triangle QRS \), the corresponding leg in \( \triangle MNP \) (let's say \( MP \)) should be equal to \( 8 \, \text{m} \)? Wait, no, wait. Wait, \( \triangle MNP \) is a right triangle with right angle at \( P \), hypotenuse \( MN = 17 \, \text{m} \), one leg \( NP = 15 \, \text{m} \). Wait, maybe we can find the other leg using Pythagorean theorem. Let the legs be \( MP \) and \( NP \), hypotenuse \( MN \). So, \( MP = \sqrt{MN^2 - NP^2} = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8 \, \text{m} \). Yes, that makes sense. So the legs of \( \triangle MNP \) are \( 8 \, \text{m} \) ( \( MP \) ) and \( 15 \, \text{m} \) ( \( NP \) ).

Step2: Calculate area of right triangle

The area of a right triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). Here, base can be \( MP = 8 \, \text{m} \) and height \( NP = 15 \, \text{m} \). So area \( = \frac{1}{2} \times 8 \times 15 \).

Step3: Compute the area

\( \frac{1}{2} \times 8 \times 15 = 4 \times 15 = 60 \, \text{m}^2 \).

Answer:

\( 60 \, \text{m}^2 \)