Question

Look at the measures shown in the adjacent figure and find the area of ⬜ PQRS.

​

Answer 1

Hey Meenag !

Answer:

690 m²

Step-by-step explanation:

Given :

PQ = 56 m

QR = 25 m

RS = 15 m

SP = 36 m

∠PSR = 90°

Formula :

Area of a triangle = $\frac{1}{2}$bh units²

• Where b is the Base.
• Where h is the Altitude.

Area of a triangle = $\sqrt{(s)(s - a)(s - b)(s - c)}$ units²

• Where a,b,c are the sides of the triangle.
• Where s is the semiperimeter of the triangle, s = $\frac{a + b + c}{2}$.

Procedure :

ar(⬜ PQRS) = ar(ΔPRS) + ar(ΔPQR)

(i) ar(ΔPRS) = (0.5) × b × h

⇒ ar(ΔPRS) = (0.5) × 15 m × 36 m

⇒ ar(ΔPRS) = (15 × 18) m²

∴ ar(ΔPRS) = 270 m².

(ii) ar(ΔPQR) = $\sqrt{(s)(s - a)(s - b)(s - c)}$ m²

PR² = PS² + SR² [Pythagoras Theorem]

⇒ PR² = (36 m)² + (15 m)²

⇒ PR² = 1296 + 225 m²

⇒ PR² = 1521 m²

∴ PR = 39 m.

Assume that

• a = PQ = 56 m.
• b = QR = 25 m.
• c = RP = 39 m.

⇒ s = $\frac{56 + 25 + 39}{2} \ m = \frac{120}{2} \ m = 60 \ m$

∴ ar(ΔPQR) = $\sqrt{(60 m)(60 m- 56 m)(60 m- 25 m)(60 m- 39m)}$

⇒ ar(ΔPQR) = $\sqrt{(60 m)(4m)(35 m)(21m)}$

⇒  ar(ΔPQR) = $\sqrt{(5 \times 12 )(2 \times 2)(5 \times 7 )(3 \times 7) \ m^{4}}$

⇒  ar(ΔPQR) = $\sqrt{(5^{2} )(2^{2})(12 \times 3 )(7^{2}) } \ m^{2}$

⇒  ar(ΔPQR) = $\sqrt{(5^{2} )(2^{2})(6^{2} )(7^{2}) } \ m^{2}$

⇒  ar(ΔPQR) = (2 × 5 × 6 × 7) m²

⇒  ar(ΔPQR) = (10 × 42) m²

∴ ar(ΔPQR) = 420 m².

Finally,

ar(⬜ PQRS) = ar(ΔPRS) + ar(ΔPQR)

⇒ ar(⬜ PQRS) = 270 m² + 420 m²

∴ ar(⬜ PQRS) = 690 m².

Thanks !

Answer 2

Answer:

690 m²

Step-by-step explanation:

Given :

PQ = 56 m

QR = 25 m

RS = 15 m

SP = 36 m

∠PSR = 90°

Formula :

Area of a triangle = \frac{1}{2}

2

1

bh units²

Where b is the Base.

Where h is the Altitude.

Area of a triangle = \sqrt{(s)(s - a)(s - b)(s - c)}

(s)(s−a)(s−b)(s−c)

units²

Where a,b,c are the sides of the triangle.

Where s is the semiperimeter of the triangle, s = \frac{a + b + c}{2}

2

a+b+c

.

Procedure :

ar(⬜ PQRS) = ar(ΔPRS) + ar(ΔPQR)

(i) ar(ΔPRS) = (0.5) × b × h

⇒ ar(ΔPRS) = (0.5) × 15 m × 36 m

⇒ ar(ΔPRS) = (15 × 18) m²

∴ ar(ΔPRS) = 270 m².

(ii) ar(ΔPQR) = \sqrt{(s)(s - a)(s - b)(s - c)}

(s)(s−a)(s−b)(s−c)

m²

PR² = PS² + SR² [Pythagoras Theorem]

⇒ PR² = (36 m)² + (15 m)²

⇒ PR² = 1296 + 225 m²

⇒ PR² = 1521 m²

∴ PR = 39 m.

Assume that

a = PQ = 56 m.

b = QR = 25 m.

c = RP = 39 m.

⇒ s = \frac{56 + 25 + 39}{2} \ m = \frac{120}{2} \ m = 60 \ m

2

56+25+39

m=

2

120

m=60 m

∴ ar(ΔPQR) = \sqrt{(60 m)(60 m- 56 m)(60 m- 25 m)(60 m- 39m)}

(60m)(60m−56m)(60m−25m)(60m−39m)

⇒ ar(ΔPQR) = \sqrt{(60 m)(4m)(35 m)(21m)}

(60m)(4m)(35m)(21m)

⇒ ar(ΔPQR) = \sqrt{(5 \times 12 )(2 \times 2)(5 \times 7 )(3 \times 7) \ m^{4}}

(5×12)(2×2)(5×7)(3×7) m

4

⇒ ar(ΔPQR) = \sqrt{(5^{2} )(2^{2})(12 \times 3 )(7^{2}) } \ m^{2}

(5

2

)(2

2

)(12×3)(7

2

)

m

2

⇒ ar(ΔPQR) = \sqrt{(5^{2} )(2^{2})(6^{2} )(7^{2}) } \ m^{2}

(5

2

)(2

2

)(6

2

)(7

2

)

m

2

⇒ ar(ΔPQR) = (2 × 5 × 6 × 7) m²

⇒ ar(ΔPQR) = (10 × 42) m²

∴ ar(ΔPQR) = 420 m².

Finally,

ar(⬜ PQRS) = ar(ΔPRS) + ar(ΔPQR)

⇒ ar(⬜ PQRS) = 270 m² + 420 m²

∴ ar(⬜ PQRS) = 690 m².

Thanks !