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Perimeter and Area - Area of a Triangle

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The area of a triangle is exactly half the area of the rectangle (or parallelogram) formed by its base and height. Any side of a triangle can be considered as the base.

A rectangle divided by a diagonal into two equal triangles, illustrating that the area of one triangle is half of the base times height.
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The height (altitude) is the perpendicular distance from the vertex to the opposite side (base). In an obtuse-angled triangle, the height may lie outside the triangle on the extension of the base.

An acute triangle showing the internal perpendicular height from the top vertex to the base.
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All triangles drawn between the same set of parallel lines and sharing the same base have equal areas, because their bases and heights are identical.

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In a right-angled triangle, the two sides containing the right angle can be taken as the base and the height respectively.

📐Formulae

Area of a triangle = 12×base×height\frac{1}{2} \times \text{base} \times \text{height}

A=12×b×hA = \frac{1}{2} \times b \times h

Base of a triangle = 2×Areaheight\frac{2 \times \text{Area}}{\text{height}}

Height of a triangle = 2×Areabase\frac{2 \times \text{Area}}{\text{base}}

💡Examples

Problem 1:

Find the area of a triangle whose base is 12 cm12\ cm and whose corresponding height is 7 cm7\ cm.

Solution:

  1. Identify the given values: base b=12 cmb = 12\ cm and height h=7 cmh = 7\ cm.
  2. Apply the formula: Area=12×b×hArea = \frac{1}{2} \times b \times h
  3. Substitute the values: Area=12×12×7Area = \frac{1}{2} \times 12 \times 7
  4. Calculate: Area=6×7=42 cm2Area = 6 \times 7 = 42\ cm^{2}.\nTherefore, the area of the triangle is 42 cm242\ cm^{2}.

Explanation:

To find the area, we simply multiply the base by the height and then divide by 22. Since both measurements are in cmcm, the final result is in cm2cm^{2}.

Problem 2:

The area of a triangle is 36 cm236\ cm^{2}. If its height is 9 cm9\ cm, find the length of its base.

Solution:

  1. Identify the given values: Area=36 cm2Area = 36\ cm^{2} and h=9 cmh = 9\ cm.
  2. Use the modified formula for base: b=2×Areahb = \frac{2 \times Area}{h}
  3. Substitute the values: b=2×369b = \frac{2 \times 36}{9}
  4. Calculate: b=729=8 cmb = \frac{72}{9} = 8\ cm.\nTherefore, the base of the triangle is 8 cm8\ cm.

Explanation:

When the area and height are known, we can rearrange the area formula to solve for the base. Multiplying the area by 22 and dividing by the height gives the required base length.

Problem 3:

Find the area of an isosceles right-angled triangle where the lengths of the two equal sides are 8 cm8\ cm each.

A right-angled triangle with both base and height labeled as 8 cm.

Solution:

Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} Area=12×8 cm×8 cm\text{Area} = \frac{1}{2} \times 8\ cm \times 8\ cm Area=12×64 cm2\text{Area} = \frac{1}{2} \times 64\ cm^2 Area=32 cm2\text{Area} = 32\ cm^2

Explanation:

In a right-angled triangle, the two sides meeting at the 90∘90^{\circ} angle serve as the base and height. Since it is isosceles, both these sides are 8 cm8\ cm.

Problem 4:

In ΔPQR\Delta PQR, the area is 20 cm220\ cm^2. If the base QRQR is 5 cm5\ cm, find the height PSPS dropped from vertex PP to the base QRQR.

Triangle PQR with base QR labeled 5 cm and altitude PS labeled as unknown height h.

Solution:

Height=2×Areabase\text{Height} = \frac{2 \times \text{Area}}{\text{base}} Height=2×20 cm25 cm\text{Height} = \frac{2 \times 20\ cm^2}{5\ cm} Height=405 cm\text{Height} = \frac{40}{5}\ cm Height=8 cm\text{Height} = 8\ cm

Explanation:

To find the height when area and base are known, we rearrange the area formula to h=2Abh = \frac{2A}{b}.