Measuring Space: Perimeter and Area | Exercise 6.2

Question 7

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

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Solution
Understand the Question
  • The diagonal PRPR of a parallelogram PQRSPQRS divides it into two triangles of equal area: Area(PSR)=Area(PQR)\text{Area}(\triangle PSR) = \text{Area}(\triangle PQR).
  • Since both PSR\triangle PSR and PQR\triangle PQR share the same base PRPR, the perpendicular heights from opposite vertices SS and QQ to the diagonal PRPR are equal.
  • The triangles PSO\triangle PSO and PQO\triangle PQO share the common base POPO and have the same perpendicular heights from SS and QQ, which means their areas are also equal.

Step 1 · Identify Common Base and Equal Altitudes

Diagram 1

In parallelogram PQRSPQRS, diagonal PRPR divides the parallelogram into two triangles of equal area: Area(PSR)=Area(PQR)\text{Area}(\triangle PSR) = \text{Area}(\triangle PQR)

Since both triangles share the base PRPR, their corresponding perpendicular heights from SS and QQ to PRPR must be equal. Let this common perpendicular height be hh.

For PSO\triangle PSO and PQO\triangle PQO:

  • Both triangles share the base POPO along the diagonal PRPR.
  • The perpendicular height of vertex SS to base POPO is hh.
  • The perpendicular height of vertex QQ to base POPO is hh.

Step 2 · Compare the Areas

Using the formula Area=12×base×height\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}:

Area(PSO)=12×PO×hArea(PQO)=12×PO×h\begin{aligned} \text{Area}(\triangle PSO) &= \dfrac{1}{2} \times PO \times h \\[0.6em] \text{Area}(\triangle PQO) &= \dfrac{1}{2} \times PO \times h \end{aligned}

Therefore, Area(PSO)=Area(PQO)\text{Area}(\triangle PSO) = \text{Area}(\triangle PQO)

Answer

Hence proved, Area(PSO)=Area(PQO)\text{Area}(\triangle PSO) = \text{Area}(\triangle PQO).

Common Mistakes
  • Assuming Congruence: Triangles PSO\triangle PSO and PQO\triangle PQO are not necessarily congruent unless OO is the midpoint of PRPR or the parallelogram is a rhombus. However, their areas are always equal.
  • Base Selection: Confusing the full diagonal PRPR with the segment POPO. While PRPR is used to establish equal altitudes, POPO is the actual base of PSO\triangle PSO and PQO\triangle PQO.

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})

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