Measuring Space: Perimeter and Area | Exercise 6.2

Question 3

Find the area of a triangle, given that its sides are 8 cm8\text{ cm} and 11 cm11\text{ cm} long, and its perimeter is 32 cm32\text{ cm}.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • To find the area of a triangle when all three sides are known, we use Heron's formula: Area=s(sa)(sb)(sc)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} where a,b,ca, b, c are the side lengths and ss is the semi-perimeter (s=a+b+c2=Perimeter2s = \dfrac{a+b+c}{2} = \dfrac{\text{Perimeter}}{2}).
  • Given two sides (8 cm8\text{ cm} and 11 cm11\text{ cm}) and the total perimeter (32 cm32\text{ cm}), we first find the unknown third side by subtracting the sum of the two given sides from the perimeter.

Step 1 · Find the Third Side

Let the two given sides be a=8 cma = 8\text{ cm} and b=11 cmb = 11\text{ cm}, and let the third side be cc.Diagram 1

c=Perimeter(a+b)=32(8+11)=3219=13 cm\begin{aligned} c &= \text{Perimeter} - (a + b) \\[0.6em] &= 32 - (8 + 11) \\[0.6em] &= 32 - 19 \\[0.6em] &= 13\text{ cm} \end{aligned}

Step 2 · Calculate the Semi-Perimeter

The semi-perimeter ss is half of the perimeter:

s=Perimeter2=322=16 cm\begin{aligned} s &= \dfrac{\text{Perimeter}}{2} \\[0.6em] &= \dfrac{32}{2} \\[0.6em] &= 16\text{ cm} \end{aligned}

Step 3 · Apply Heron's Formula

Using Heron's formula:

Area=s(sa)(sb)(sc)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}

Substitute s=16s = 16, a=8a = 8, b=11b = 11, and c=13c = 13:

Area=16(168)(1611)(1613)=16×8×5×3=1920=64×30=830 cm2\begin{aligned} \text{Area} &= \sqrt{16(16-8)(16-11)(16-13)} \\[0.6em] &= \sqrt{16 \times 8 \times 5 \times 3} \\[0.6em] &= \sqrt{1920} \\[0.6em] &= \sqrt{64 \times 30} \\[0.6em] &= 8\sqrt{30}\text{ cm}^2 \end{aligned}
Answer

830 cm28\sqrt{30}\text{ cm}^2

Common Mistakes
  • Using Perimeter instead of Semi-Perimeter: Substituting s=32s = 32 instead of s=16s = 16 into Heron's formula.
  • Radical Simplification Error: Factoring incorrectly under the square root; simplify as 16×8×5×3=16×4×2×15=64×30=830\sqrt{16 \times 8 \times 5 \times 3} = \sqrt{16 \times 4 \times 2 \times 15} = \sqrt{64 \times 30} = 8\sqrt{30}.
  • Unit Omission: Forgetting that area must be expressed in square units (cm2\text{cm}^2) rather than linear units (cm\text{cm}).

More questions in Exercise 6.2

Q1

Find the area of triangle ADE in Fig. 6.31.

Q2

The parallel sides of a trapezium are 40 cm40\text{ cm} and 20 cm20\text{ cm}. If its non-parallel sides are both equal, each being 26 cm26\text{ cm}, find the area of the trapezium.

Q3

Find the area of a triangle, given that its sides are 8 cm8\text{ cm} and 11 cm11\text{ cm} long, and its perimeter is 32 cm32\text{ cm}.

Q4

The sides of a triangular plot are in the ratio 3:5:73 : 5 : 7; its perimeter is 300 m300\text{ m}. Find its area.

Q5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2128\text{ cm}^2, find the length of the shorter diagonal.

Q6

ABCDABCD is a parallelogram. PP and QQ are any two points on side ABAB. What can you say about the ratio area(ΔPCD):area(ΔQCD)\text{area}(\Delta \text{PCD}) : \text{area}(\Delta \text{QCD})?

Q7

OO is any point on the diagonal PRPR of a parallelogram PQRSPQRS. Prove that the areas of triangles PSOPSO and PQOPQO are equal.

Q8

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

Q9

In ΔABC\Delta\text{ABC}, the midpoint of BC\text{BC} is D\text{D} (Fig. 6.32). Median AD\text{AD} is drawn. P\text{P} is any point on AD\text{AD}. Show that area(ΔABP)=area(ΔACP)\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP}).

Q10

Given a square ABCDABCD, let PP be a point within it. Join PAPA, PBPB, PCPC, PDPD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB\Delta PAB and ΔPCD\Delta PCD) and the green region (ΔPBC\Delta PBC and ΔPDA\Delta PDA)?

Q11

In ΔABC\Delta\text{ABC}, DD is the midpoint of AB\text{AB}. PP is any point on BC\text{BC}, and QQ is a point on AB\text{AB} such that CQPD\text{CQ} \parallel \text{PD}. PQ\text{PQ} is joined (Fig. 6.34). Prove that

Area(ΔBPQ)=12Area(ΔABC)\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})

← Back to Measuring Space: Perimeter and Area