Area of parallelograms and triangles

You can find the area of a rec­tan­gle by mul­ti­ply­ing its length and breadth. But most shapes are not rec­tan­gles. In this les­son you will see that a par­al­lel­o­gram can be cut and rearranged into a rec­tan­gle, and that a tri­an­gle is exactly half of a par­al­lel­o­gram. That gives two short for­mu­las, both built on the idea of a base and a height.

Base and height#

Choose one side of a shape and call it the base. The height is the per­pen­dic­u­lar dis­tance from the base to the oppo­site side or ver­tex. Per­pen­dic­u­lar means at a right angle.

The height is not the slant­ing side. It is the straight-up dis­tance, like the height of a per­son stand­ing on the floor, mea­sured at a right angle to the base.

Area of a par­al­lel­o­gram#

Take a par­al­lel­o­gram drawn on paper. From one end of the top side, draw a line straight down to the base, at a right angle. This cuts off a right-angled tri­an­gle at one end. Cut it off and slide it to the other end. It fits exactly, and the par­al­lel­o­gram becomes a rec­tan­gle.

The rec­tan­gle's length is the base of the par­al­lel­o­gram, and its breadth is the height. Noth­ing was added or lost, so

$$\text{area of a parallelogram} = \text{base} \times \text{height} = b \times h$$

Exam­ple 1#

A par­al­lel­o­gram has base 12 cm and height 7 cm. Find its area.

$$12 \times 7 = 84$$

The area is 84 cm².

Exam­ple 2#

The area of a par­al­lel­o­gram is 96 cm² and its base is 12 cm. Find its height.

$$h = 96 \div 12 = 8$$

The height is 8 cm.

Exam­ple 3: Either side can be the base#

The sides of a par­al­lel­o­gram are 15 cm and 10 cm. The height on the 15 cm side is 6 cm. Find the height on the 10 cm side.

Area $= 15 \times 6 = 90$ cm².

The area is the same whichever side we use as the base. So the height on the 10 cm side is $90 \div 10 = 9$ cm.

Area of a tri­an­gle#

Take two copies of the same tri­an­gle. Turn one of them round and place it against the other along a side. Together they make a par­al­lel­o­gram with the same base and the same height as the tri­an­gle. So the tri­an­gle is half of that par­al­lel­o­gram:

$$\displaystyle \text{area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}bh$$

Any side can be the base. The height is then the per­pen­dic­u­lar from the oppo­site ver­tex to that base. In an obtuse-angled tri­an­gle, the height on a short side falls out­side the tri­an­gle, on the base drawn longer. The for­mula still works.

Exam­ple 4#

Find the area of a tri­an­gle with base 10 cm and height 6 cm.

$$\displaystyle \frac{1}{2} \times 10 \times 6 = 30$$

The area is 30 cm².

Exam­ple 5#

The area of a tri­an­gle is 54 cm² and its base is 12 cm. Find its height.

$$\displaystyle \frac{1}{2} \times 12 \times h = 54 \quad \text{so} \quad 6h = 54 \quad \text{and} \quad h = 9$$

The height is 9 cm.

Right-angled tri­an­gles#

In a right-angled tri­an­gle, the two sides that make the right angle are per­pen­dic­u­lar to each other. So one can be the base and the other the height.

$$\displaystyle \text{area} = \frac{1}{2} \times (\text{one side of the right angle}) \times (\text{the other})$$

Exam­ple 6#

A right-angled tri­an­gle has sides 6 cm, 8 cm and 10 cm. The right angle is between the 6 cm and 8 cm sides. Find its area, and the height on the 10 cm side.

$$\displaystyle \text{area} = \frac{1}{2} \times 6 \times 8 = 24$$

The area is 24 cm².

Now use the 10 cm side as the base. The area is still 24 cm²:

$$\displaystyle \frac{1}{2} \times 10 \times h = 24 \quad \text{so} \quad 5h = 24 \quad \text{and} \quad h = 4.8$$

The height on the 10 cm side is 4.8 cm.

Exam­ple 7: Half of a rec­tan­gle#

A rec­tan­gle 14 cm by 9 cm is cut along a diag­o­nal. Find the area of each tri­an­gle.

The diag­o­nal cuts the rec­tan­gle into two equal right-angled tri­an­gles. Each has area

$$\displaystyle \frac{1}{2} \times 14 \times 9 = 63$$

so 63 cm², which is half of the rec­tan­gle's 126 cm².

Exam­ple 8: A tri­an­gle inside a par­al­lel­o­gram#

A tri­an­gle and a par­al­lel­o­gram stand on the same base of 16 cm and have the same height of 5 cm. Com­pare their areas.

Par­al­lel­o­gram: $16 \times 5 = 80$ cm². Tri­an­gle: $\displaystyle \frac{1}{2} \times 16 \times 5 = 40$ cm².

The tri­an­gle has exactly half the area of the par­al­lel­o­gram. This is true when­ever a tri­an­gle and a par­al­lel­o­gram have the same base and the same height.

Prac­tice#

1. Find the area of a par­al­lel­o­gram with base 9 cm and height 6.5 cm.
2. The area of a par­al­lel­o­gram is 135 m² and its height is 9 m. Find its base.
3. The sides of a par­al­lel­o­gram are 20 cm and 12 cm. The height on the 20 cm side is 9 cm. Find the height on the 12 cm side.
4. Find the area of a tri­an­gle with base 15 cm and height 8 cm.
5. The area of a tri­an­gle is 72 cm² and its height is 16 cm. Find its base.
6. A right-angled tri­an­gle has the sides of its right angle 12 cm and 9 cm. Find its area.
7. A right-angled tri­an­gle has sides 5 cm, 12 cm and 13 cm, with the right angle between the 5 cm and 12 cm sides. Find the height on the 13 cm side.
8. A square of side 10 cm is cut along a diag­o­nal. Find the area of each tri­an­gle.
9. A tri­an­gu­lar field has base 120 m and height 85 m. Find the cost of lev­el­ling it at ₹4 per square metre.

Answers#