## Solid Geometry Chapter 6 Construction Drawing Of Triangles Equal To The Area Of A Given Quadrilateral

**Chapter 6 Construction Drawing Of Triangles Equal To The Area Of A Given Quadrilateral ****Construction of a triangle of area equal to the area of a given quadrilateral**

In the previous chapter we learned how to construct a parallelogram of area equal to the area of a given triangle and there we have taken the help of the theorem, the area of a triangle is half of the area of a parallelogram having the same base and within the same parallels.

Again, we know that the areas of all the triangles having the same base and within the same parallels are equal in area.

We shall hence learn how to construct a triangle of area equal to the area of a given quadrilateral with the help of this theorem.

**Read and Learn More WBBSE Solutions For Class 9 Maths**

**Construction – 1**

**Construction of triangles of area equal to the area of a given quadrilateral.**

Let ABCD is a given quadrilateral.

We have to construct a triangle of area equal to the area of ABCD.

**Method of construction:**

1. Let us first draw the diagonal DB.

2. Let us now draw an arc on the produced side of AB with C as the center and with a radius equal to DB and let us draw another arc on the same side with B as the center and with a radius equal to DC.

Let these two arcs intersect each other at P.

Then CP || DB is constructed.

3. Let the produced AB intersects CP at E.

4. D and E are joined.

Then ΔADE is the required triangle.

**Proof:**

Δ BCD and ΔBDE have the same base BD and they lie within the same parallels BD and EC [∵ by construction BD || EC ].

∴ Δ BCD = ΔBDE

or, Δ BCD + Δ ABD = ΔBDE + Δ ABD [ by adding ΔABD to both sides]

or, quadrilateral ABCD = triangle ADE.

∴ area of the quadrilateral ABCD = area of the ΔADE.

Hence, ΔADE is the required triangle. (Proved)

**Application-1: Construction of a rectangle of area equal to the area of a given quadrilateral**

Let ABCD is a given quadrilateral. We have to construct a rectangle of area equal to the area of ABCD.

**Method of construction :**

1. Let us draw the diagonal DB of ABCD.

2. Let us draw CG through C parallel to DB.

3. Let us produce AB, which intersects CG at Q.

4. D and Q are joined. Thus a triangle Δ ADQ is produced.

5. Let us draw EF through D and parallel to AQ.

6. Let us draw PS, the perpendicular – bisector of AQ, which intersect AQ at P and EF at S.

7. Let us cut the part SR from SF equal to PQ.

8. R and Q are joined. Then PQRS is the required rectangle.

**Proof:** D and P are joined.

Δ BCD and Δ BDQ have the same base DB and they lie within the same parallels DB and CG.

∴ Δ BCD = Δ BDQ

or, Δ BCD + Δ ABD = Δ BDQ+Δ ABD [ adding A ABD to both sides]

or, quadrilateral ABCD = triangle ADQ………. (1)

Again, P is the mid-point of AQ,

∴ DP is a median of Δ ADQ.

∴ Δ DPQ = \(\frac{1}{2}\)Δ ADO

Now, in PQRS, PQ = SR and PQ || SR and SPQ 90° [∵ SP ⊥ AQ ]

∴ PQRS is a rectangle.

Again, Δ DPQ and PQRS have the same base PQ and they lie within the same parallels PQ

and EF.

Δ DPQ = \(\frac{1}{2}\) PQRS ……………….(3)

Then from (2) and (3) we get, \(\frac{1}{2}\) Δ ADQ – \(\frac{1}{2}\) PQRS

or, Δ ADQ = PQRS…… (4)

Again, from (1) and (4) we get, quadrilateral ABCD = PQRS.

∴ PQRS is the required rectangle. (Proved)

**Application-2: Construct a parallelogram of an area equal to the area of a given quadrilateral, one of whose angles is a given fixed angle.**

Let ABCD be a given quadrilateral and O be a given fixed angle.

We have to construct a parallelogram of an area equal to the area of ABCD and one of whose angles is equal to 0.

**Method of construction:**

1. Let us draw the diagonal DB of ABCD.

2. Let us draw DB || CG through C.

3. Let us produce AB which intersects CG at Q.

4. D and Q are joined. Then A BDQ is produced.

5. Let us now draw EF through D parallel to AQ.

6. Let us determine the mid-point P of AQ.

7. Let us then draw QPS at P equal to 0. Let the side PS of QPS intersects EF and S.

8. Let us now cut the part SR from SF equal to PQ.

9. R and Q are joined.

10. Then PQRS is the required parallelogram.

## Chapter 6 Construction Drawing Of Triangles Equal To The Area Of A Given Quadrilateral Construction Of A Triangle And A Quadrilateral Of Area Equal To The Area Of A Given Pentagon

Using the theorem that the area of all triangles having the same base (or equal base) and within the same parallels are equal in area, we shall now construct a triangle and a quadrilateral of area equal to the area of a given pentagon.

**CONSTRUCTION-2**

Construction of

1. A triangle and

2. A quadrilateral of area equal to the area of a given Pentagon.

Let ABCDE be a given pentagon.

We have to construct

1. A triangle and

2. A quadrilateral of area equal to the area of a ABCDE.

**Method of construction :**

1. Let us draw the diagonals AC and AD of the pentagon ABCDE.

2. Let us draw BF through B parallel to AC.

3. Let us now draw EG through E parallel to AD.

4. Let us produce DC on the left side which intersects BF at P and CD on the right side which intersects EG at Q.

5. A, P, and A, Q are joined.

Then,

1. ΔAPQ is the required triangle

2. ΔPDE G is the required quadrilateral.

Proof:

1. ΔABC and Δ APC have the same base AC and they lie within the same parallels AC and BF ( by construction AC || BF),

Δ ABC = Δ APC…..…. (1)

Again, Δ AED and ΔAQD have the same base AD and they lie within the same parallels AD and EG (by construction AD || EG),

∴ ΔAED = ΔAQD……………..(2)

Now, adding (1) and (2) we get, ΔABC + ΔAED = ΔAPC + ΔAQD

or, ΔABC + ΔACD + ΔAED = ΔAPC+Δ ACD + ΔAQD……….(3) (adding ΔACD to both the sides)

or, Pentagon ABCDE = ΔAPQ.

Area of the pentagon ABCDE = area of the ΔAPQ.

∴APQ is the required triangle. (Proved)

2. Again, from (3) we get, ΔABC +ΔACD + ΔAED = ΔAPC+ΔACD + ΔAQD

or, Pentagon ABCDE = ΔAPC + ΔACD + ΔAED [ by (2)]

∴ Pentagon ABCDE = quadrilateral APDE.

∴ Area of the pentagon ABCDE = area of the quadrilateral APDE.

∴ APDE is the required quadrilateral. (Proved)

## Chapter 6 Construction Drawing Of Triangles Equal To The Area Of A Given Quadrilateral In The Following Examples How These Constructions Are Applied To Real Problems Is Discussed Thoroughly

**Question 1. Construct a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm, and ABC = 60″. Then construct a triangle of area equal to the area of that quadrilateral. (Give method, signs, and proof of construction)**

**Solution: **

**Given**

**A Quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm, and ABC = 60°. **

Let AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm and ABC = 60°. We have to construct.

1. the quadrilateral ABCD.

2. a triangle of area equal to the area of ABCD.

**Method of construction :**

1. Let us draw any straight line BX.

2. Let us cut the part BC from BX equal to 6 cm.

3. Let us draw ∠CBY = 60° at B along the side BC.

4. Let us cut part AB from the side BY of CBY.

5. Let us draw two arcs on the interior region of ∠ABC, one with center A and a radius 3 cm and the other with center C and a radius 4 cm.

Let these two arcs intersect each other at D.

6. A, D, and C, D are joined.

The quadrilateral ABCD is constructed.

7. Now, A and C are joined.

8. Let us then draw DF through D parallel to AC.

9. Let us produce BC, which intersects DF at E.

10. A and E are joined.

Then, ΔABE is the required triangle to be constructed.

**Proof :** By construction, in the quadrilateral ABCD, AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm and ∠ABC = 60°

∴ ABCD is the required quadrilateral.

## Chapter 6 Construction Drawing Of Triangles Equal To The Area Of A Given Quadrilateral In The Following Examples, How These Constructions Are Applied To Real Problems Is Discussed Thoroughly

**Question 1. Construct a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm, and ABC = 60″. Then construct a triangle of area equal to the area of that quadrilateral. (Give method, signs, and proof of construction)**

**Solution: **

**Given **

**A Quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm, and ABC = 60°.**

Let AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm and ABC = 60°.

We have to construct.

1. The quadrilateral ABCD.

2. A triangle of area equal to the area of ABCD.

**Method of construction :**

1. Let us draw any straight line BX.

2. Let us cut the part BC from BX equal to 6 cm.

3. Let us draw ∠CBY = 60° at B along the side BC.

4. Let us cut part AB from the side BY of CBY.

5. Let us draw two arcs on the interior region of ∠ABC, one with center A and a radius of 3 cm and the other with center C and a radius of 4 cm.

Let these two arcs intersect each other at D.

6. A, D, and C, D are joined.

The quadrilateral ABCD is constructed.

7. Now, A and C are joined.

8. Let us then draw DF through D parallel to AC.

9. Let us produce BC, which intersects DF at E.

10. A and E are joined.

Then, ΔABE is the required triangle to be constructed.

**Proof :** By construction, in the quadrilateral ABCD, AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm and ∠ABC = 60°

∴ ABCD is the required quadrilateral.

Again, Δ ACD and Δ ACE have the same base AC and they lie within the same parallels AC and DF.

∴ Δ ACD = Δ ACE

or, Δ ACD + Δ ABC =Δ ACE + Δ ABC (adding Δ ABC to both sides.)

or, quadrilateral ABCD = Δ ABE area of the quadrilateral ABCD = area of the Δ ABE.

∴ Δ ABE is the required triangle. (Proved)

**Question 2.** **Construct a quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 3.7 cm, and diagonal AC = 6 cm. Then construct a triangle of area equal to the area of ABCD. (Give method, signs, and proof of constructions.)**

**Solution: **

**Given **

**A Quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 3.7 cm, and diagonal AC = 6 cm. **

Let in the quadrilateral ABCD, AB = 4 cm, BC= 5 cm, CD = 4.8 cm, DA = 3.7 cm and diagonal AC = 6 cm.

We have to construct the quadrilateral ABCD and then a triangle of area equal to the area of ABCD.

**Method of construction :**

1. Let us construct the quadrilateral ABCD of which AB = 4 cm, BC= 5 cm, CD = 4.8 cm, DA = 37 cm and diagonal AC = 6 cm.

2. Let us draw DG through D parallel to AC.

3. Let us produce BC which intersects DG at E.

4. A and E are joined.

∴ ΔABE is the required triangle.

**Proof:**

By construction, in the quadrilateral ABCD, AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 3.7 cm, and diagonal AC = 6 cm.

∴ ABCD is the required quadrilateral.

Again, Δ ACD and ΔACE have the same base AC and they lie within the same parallels AC and DG.

∴ ΔACD =ΔACE

or, ΔACD +Δ ABC = ΔACE + ΔABC (adding Δ ABC to both sides)

or, quadrilateral ABCD = ΔABE.

ΔABE is the required triangle. (Proved)

**Question 3.** **Construct a quadrilateral ABCD of which AB = 4 cm, BC = 6 cm, CD = 3 cm, ∠ABC = 60**°**, 2 BCD = 55**°**. Then construct a triangle of area equal to the area of ABCD and of which one of the sides remains on AB and the other one remains on BC. (Give construction signs only)**

**Solution:** Here, ΔABE is the required triangle one of whose sides is on AB and the other on BC.

**Given** **A Quadrilateral ABCD of which AB = 4 cm, BC = 6 cm, CD = 3 cm, ∠ABC = 60**°**, 2 BCD = 55**°**.**

**Ex. 4. Construct a square of sides 6 cm and then construct a triangle of area equal to the area of this square. (Method, signs, and proof of construction must have to be given.)**

**Solution:**

Let ABCD be a square each of whose side is 6 cm.

We have to construct this square ABCD.

Then we have to construct a triangle of area equal to the area of ABCD.

**Method of construction :**

**1. Construction of the square:**

1. Let us first draw a line segment AB equal to 6 cm.

2. Let us draw ∠BAX = 90° at A of AB.

3. Let us cut the part AD from the side AX of BAX equal to 6 cm.

4. Let us now draw two arcs on the interior region of BAD, one with center B and the other with center D, and both with a radius 6 cm.

Let these two arcs intersect each other at C.

5. B, C, and D, C are joined.

Then, ABCD is the required square.

**2.**** Construction of triangle:**

1. Let us draw the diagonal DB.

2. Let us draw CG through C parallel to DB.

3. Let us produce AB, which intersects CG at E.

4. Then A ADE is the required triangle.

Proof By construction, in the quadrilateral ABCD, AB = BC = CD = DA = 6 cm and < BAD = 90°, i.e., all the 4 sides of ABCD are equal and one of its angles is 90°.

ABCD is the required square.

Again, Δ BCD and Δ BDE have the same base DB and they lie within the same parallels DB and CG.

∴ ΔBCD = Δ BDE

or, ΔBCD + ΔABD =Δ BDE + ΔABD (adding Δ ABD to both sides)

or, Square ABCD = ΔADE.

Hence, ΔADE is the required triangle. (Proved)

**Question 5.** **Construct a quadrilateral ABCD of which AD ⊥ ****AB, BC ⊥ AB, and AB = 5 cm, ****AD = 7 cm, and BC = 4 cm. Then construct a triangle of area equal to the area of ABCD and one ****of whose angles is 30°. (Give construction signs only)**

**Construction:**

**Given **

**A Quadrilateral ABCD of which AD ⊥ ****AB, BC ⊥ AB, and AB = 5 cm, ****AD = 7 cm, and BC = 4 cm. **

Here, ΔAEF is the required triangle of which ∠EAF = 30%, i.e., quadrilateral ABCD = ΔADE = Δ ΑΕΕ.

**Question 6.** **Construct any pentagon ABCDE and then construct a triangle of area equal to the area of ABCDE and one of whose vertices is C. (Give method, signs, and proof of construction.) **

**Solution:**

Let ABCDE be a pentagon.

We have to construct a triangle of area equal to the area of ABCDE and one of whose vertices is C.

**Method of construction :**

1. Let us draw the diagonals CE and CA of the pentagon ABCDE.

2. Let us now draw DG through D parallel to CE.

3. Let us then draw BF through B parallel to CA.

4. Let us produce AE (on the left side), which intersects DG at P.

5. Let us produce EA (on the right side), which intersects BF at Q.

6. C, P and C, Q are joined.

Then, ΔPCQ is the required triangle to be constructed.

**Proof:** ΔCDE and ΔCEP have the same base CE and they lie within the same parallels CE and DG. (∵ by construction CE || DG),

∴ ΔCDE = ΔCEP……….. (1)

Again, Δ ABC and ΔACQ have the same base CA and they lie within the same parallels CA and BF. (∵ by construction CA || BF),

∴ Δ ABC = ΔACQ………….(2)

Now, (1) and (2) we get, ΔCDE + ΔABC = ΔCEP + ΔACQ

or, ΔCDE + ΔACE +ΔABC = ΔCEP + ΔACE + ΔACQ (adding ΔACE to both sides)

or, pentagon ABCDE = ΔPCQ.

∴ area of the pentagon ABCDE = area of the ΔPCQ.

∴ ΔPCQ is the required triangle. (Proved)