Go through the **Spectrum Math Grade 6 Answer Key Chapter 6 Lesson 6.1 Calculating Area: Triangle** and get the proper assistance needed during your homework.

## Spectrum Math Grade 6 Chapter 6 Lesson 6.1 Calculating Area: Triangle Answers Key

The area (A) of a triangle is one-half the of the base (b) times the height (h).

A = \(\frac{1}{2}\) × 9 × 4

= \(\frac{1}{2}\) × 36

= 18

A = 18 square inches

A = \(\frac{1}{2}\) × b × h

or

A = \(\frac{1}{2}\)bh

A = \(\frac{1}{2}\) × 5 × 7

= \(\frac{1}{2}\) × 35

= 17\(\frac{1}{2}\)

A = 17\(\frac{1}{2}\) square inches

**Find the area of each right triangle.**

Question 1.

a.

A = ____ sq. in.

Answer:

The base length of given triangle is 8 in. and the perpendicular line from the base to the opposite vertex is the height which is 10 in. long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 8 × 10

= \(\frac{1}{2}\) × 80

= 40 sq.in

Thus, the area of triangle will be 40 sq.in

b.

A = ____ sq. ft.

Answer:

The base length of given triangle is 11 ft and the perpendicular line from the base to the opposite vertex is the height which is 5 ft long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 11 × 5

= \(\frac{1}{2}\) × 55

= 27.5 sq.ft

Thus, the area of triangle will be 27.5 sq.ft

Question 2.

a.

A = ____ sq. ft.

Answer:

The base length of given triangle is 5 ft and the perpendicular line from the base to the opposite vertex is the height which is 5 ft long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 5 × 5

= \(\frac{1}{2}\) × 25

= 12.5 sq.ft

Thus, the area of triangle will be 12.5 sq.ft

b.

A = ____ sq. yd.

Answer:

The base length of given triangle is 12 yd and the perpendicular line from the base to the opposite vertex is the height which is 6 yd long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 12 × 6

= \(\frac{1}{2}\) × 72

= 36 sq.yd

Thus, the area of triangle will be 36 sq.yd

The area of a triangle is related to the area of a rectangle.

The dashed line indicates the height of the triangle, rectangle: A = 8 × 6 = 48 sq. units triangle: A = \(\frac{1}{2}\)(8) (6) = 24 sq. units

A = \(\frac{1}{2}\)(4.5)(3) = 6\(\frac{3}{4}\)sq. m

Notice that in a right triangle the height is the length of one of the legs. This is not the case with acute and obtuse triangles.

**Find the area of each triangle below.**

Question 1.

a.

A = ____ sq. ft.

Answer:

The base length of given triangle is 11 ft and the dashed line is the height which is 5 ft long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 11 × 5

= \(\frac{1}{2}\) × 55

= 27.5 sq.ft

Thus, the area of triangle will be 27.5 sq.ft

b.

A = ____ sq. yd.

Answer:

The base length of given triangle is 12 yd and the perpendicular line from the base to the opposite vertex is the height which is 8 yd long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 12 × 8

= \(\frac{1}{2}\) × 96

= 48 sq.yd

Thus, the area of triangle will be 48 sq.yd

c.

A = ____ sq. in.

Answer:

The base length of given triangle is 19 in. and the dashed line is the height which is 11 in. long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 19 × 11

= \(\frac{1}{2}\) × 209

= 104.5 sq.in.

Thus, the area of triangle will be 104.5 sq.in.

Question 2.

a.

A = ____ sq. ft.

Answer:

The base length of given triangle is 5 ft and the perpendicular line from the base to the opposite vertex is the height which is 4 ft long

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 5 × 4

= \(\frac{1}{2}\) × 20

= 10 sq.ft

Thus, the area of triangle will be 10 sq.ft

b.

A = ____ sq. cm.

Answer:

The base length of given triangle is 17 cm and the dashed line is the height which is 14.5 cm long.

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 17 × 14.5

= \(\frac{1}{2}\) × 246.5

= 123.25 sq.cm

Thus, the area of triangle will be 123.25 sq.cm

c.

A = ____ sq. m.

Answer:

The base length of given triangle is 8 m and the dashed line is the height which is 8 m long.

A = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × 8 × 8

= \(\frac{1}{2}\) × 64

= 32 sq.m

Thus, the area of triangle will be 32 sq.m

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