Exercises: Derive triangle area formula - Common Core Math - HS Geometry

A

Recall / Warm-Up

1.

A triangle has a base of 8 cm and a perpendicular height of 5 cm. What is its area?

A.

40 cm²

B.

20 cm²

C.

13 cm²

D.

80 cm²

2.

In a right triangle, an acute angle is $\theta$ . The side opposite $\theta$ has length $h$ and the hypotenuse has length $a$ . Which equation is correct?

A.

$\sin(\theta) = \dfrac{a}{h}$

B.

$\sin(\theta) = \dfrac{h}{a}$

C.

$\sin(\theta) = \dfrac{h}{a^2}$

D.

$\cos(\theta) = \dfrac{h}{a}$

3.

An altitude of a triangle is a line segment drawn from a vertex that is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ .

A.

parallel to the opposite side

B.

perpendicular to the opposite side (or its extension)

C.

equal in length to the opposite side

D.

bisecting the opposite side

B

Fluency Practice

1.

In the derivation of $A = \dfrac{1}{2}ab\sin(C)$ , an altitude $h$ is drawn from vertex $B$ perpendicular to side $b$ (= $AC$ ). In the right triangle formed at vertex $C$ , the sine definition gives $\sin(C) = \dfrac{h}{a}$ . Solving for $h$ : $h =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ . Substituting into $A = \dfrac{1}{2} \cdot b \cdot h$ gives $A =$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ .

altitude h:

area formula:

2.

Find the area of a triangle with sides $a = 5$ cm and $b = 7$ cm and included angle $C = 60\degree$ . Use $\sin(60\degree) \approx 0.866$ . Round to the nearest tenth.

$Triangle PQR with sides PQ = 10, PR = 12, and included angle P = 50 degrees, with the area labeled as unknown$

3.

Triangle $PQR$ has sides $PQ = 10$ and $PR = 12$ , with included angle $P = 50\degree$ . Find the area of triangle $PQR$ . Use $\sin(50\degree) \approx 0.766$ . Round to the nearest tenth.

4.

A triangle has sides $a = 6$ m and $b = 8$ m with included angle $C = 120\degree$ . Find its area. Use $\sin(120\degree) \approx 0.866$ . Round to the nearest tenth.

5.

Find the area of a triangle with two sides of length 4 ft and 9 ft and an included angle of 45°. Use $\sin(45\degree) \approx 0.707$ . Round to the nearest tenth.

C

Varied Practice

D

Word Problems

1.

A triangular garden has two sides of 12 m and 15 m with an included angle of 70°.

Find the area of the garden. Use $\sin(70\degree) \approx 0.940$ . Round to the nearest tenth of a square metre.

$Top-down diagram of a triangular land plot with two sides of 50 m and 70 m, an included angle of 110 degrees, and the area labeled as unknown$

2.

A surveyor measures a triangular land plot and records two boundary sides as 50 m and 70 m with an included angle of 110° between them.

Find the area of the land plot. Use $\sin(110\degree) \approx 0.940$ . Round to the nearest whole square metre.

$Parallelogram with sides 8 cm and 6 cm, included angle 70 degrees, divided by a diagonal into two congruent triangles$

3.

A parallelogram has adjacent sides of 8 cm and 6 cm with an included angle of 70°. A diagonal divides it into two congruent triangles.

1.

Find the area of one triangle formed by the diagonal. Use $\sin(70\degree) \approx 0.940$ . Round to the nearest tenth.

2.

Find the total area of the parallelogram. Round to the nearest tenth.

E

Error Analysis

1.

Triangle $XYZ$ has $XY = 10$ , $YZ = 12$ , and angle $X = 50\degree$ . Priya computes:
$A = \frac{1}{2}(10)(12)\sin(50\degree) \approx \frac{1}{2}(120)(0.766) \approx 45.96$

What error, if any, did Priya make?

A.

Priya used angle $X$ , but angle $X$ is not the included angle between sides $XY$ and $YZ$ .

B.

Priya should have used $\cos(50\degree)$ instead of $\sin(50\degree)$ .

C.

Priya forgot the factor of 1/2.

D.

Priya's work is correct.

2.

Jamal computes the area of a triangle with sides 5 and 8 and included angle 30°:
$A = (5)(8)\sin(30\degree) = 40 \times 0.5 = 20$

What mistake did Jamal make?

A.

Jamal used the wrong trig function. He should have used $\cos(30\degree)$ .

B.

Jamal omitted the factor of $\dfrac{1}{2}$ . The correct area is $\dfrac{1}{2}(5)(8)\sin(30\degree) = 10$ .

C.

Jamal used a non-included angle.

D.

Jamal should have used the Law of Cosines instead.

F

Challenge / Extension

1.

A triangular solar panel has two sides of 25 m and 30 m with an included angle of 55°. Find the area of the panel. Use $\sin(55\degree) \approx 0.819$ . Round to the nearest tenth of a square metre.

2.