Selina Concise Class 9 Maths Chapter 16 Area Theorems Exercise 16C Solutions
EXERCISE – 16C
(Q1) In the given figure, the diagonals AC and BD intersect at point O if OB = OD and AB||DC, show that:
(i) Area (△DOC) = Area (△AOB)
(ii) Area (△DCB) = Area (△ACB)
(iii) ABCD is a parallelogram
Solution:
Given: OB = OD and AB = DC
To Prove:
(i) Area (△DOC) = Area (△AOB)
(ii) Area (△DCB) = area (△ACB)
(iii) ABCD is a parallelogram
Proof:
In △DOC and △AOB,
∠DOC = ∠AOB (vertically opposite angles)
OD = OB (∵ Given)
∠CDO = ∠OBA (Alternate interior angles)
∴ △DOC ≅ △AOB (By angle-side-angle (ASA)
Then, area (△DOC) = area (△AOB) —– (i)
Now, area (△DCB) = area (△DOC) + area (△BOC)
= area (△AOB) + area (△BOC)
(ii) area (△DCB) = area (△ABC)
(iii) (△DCB) and (△ABC) are lie in same base and equal in area. They lie between parallel lines.
If AB parallel DC (AB||DC) and AD parallel BC (AD||BC)
Hence, ABCD is a parallelogram.
(Q2) The given figure shows a parallelogram ABCD with area 324 sq. cm P is a point in AB such that AP:PB = 1:2 Find the area of △APD
Given:
Area (ABCD) = 324 cm2
AP : PB = 1:2
Find:- area (△APD) =?
Solution:
Area (△ADB) = area (△BCD)
Area (△ADB) = 1/2 area (ABCD)
= 1/2 × 324
= 162 cm2
Now, area (△APD)/area (△ADB) = x/3x
Area (△APD) = 1/3 area (△ADB)
= 1/3 × 162
= 54 cm2
(Q3) In △ABC, E and F are mid-points of sides AB and AC respectively. If BF and CE intersect each other at point O, Prove that the △OBC and quadrilateral AEOF are equal in area.
Solution:
Given: E and F are mid points of AB and AC
To Prove: area (△BOC) = area (AEOF)
Proof: area (△BEC) = area (△BFC)
Area (△BEC) – area (△BOC) = area (△BFC) – area (△BOC)
Area (△BOE) = area (△COF) —- (i)
In △ABC,
Area (△BAF) = area (△BCF) —– (ii) (∵ BF is a median)
Area (△BAF) – area (△BOE) = area (△BCF) – area (△COF)
Area (AEOF) = area (△BOC) (hence Proved)
(Q4) In parallelogram ABCD, P is mid-point of AB. CP and BD intersect each other at point O. If area of △POB = 40cm2, and OP : OC = 1:2, find
(i) Areas of △BOC and △PBC
(ii) Areas of △ABC and parallelogram ABCD
Given: area (△POB) = 40cm2
OP : OC = 1:2
Find :
(i) Areas of △BOC and △PBC
(ii) Area (△ABC) and parallelogram ABCD
Solution:
In △PBC,
Area(△POB)/area(△BOC) = 1x/2x
40/area (△BOC) = 1/2
40×2 = area (△BOC)
80 cm2 = area (△BOC)
Area (△PBC) = area (△POB) + area (△BOC)
= 40 + 80
= 120 cm2
(ii) area (△ACD) = area (△PBC) = 120 cm2
Area (△ABC) = area (△ACP) + area (△PBC)
= 120 + 120
= 240 cm2
Now, area (ABCD) = 2 area (△ABC)
= 2×240
= 480 cm2
(Q5) The medians of a triangle ABC intersect each other at point G. If one of it’s medians is AD.
Prove that:
(i) Area (△ABD) = 3 × area (△BGD)
(ii) Area (△ACD) = 3 × area (△CGD)
(iii) Area (△BGC) = —- × area (△ABC)
Solution:
Given: The medians of a triangle ABC intersect each other at point G
To Prove:
(i) area (△ABD) = 3×area (△BGD)
(ii) area (△ACD) = 3 area (△CGD)
(iii) area (△BGC) = 1/3 area (△ABC)
Proof:
Area (△BGD)/area (△BGA) = 1x/2x
Area (△BGD) = 1/2 area (△BGA)
2 area (△BGD) = area (△BGA) —– (i)
Now, Area (△ABD) = area (△BGD) + area (△BGA)
= area (△BGD) + 2 area (△BGD) (Using equating (i))
Area (△ABD) = 3 area (△BAD) —- (ii) (Hence Proved)
Similarly, area (△ACD) = 3 area (△CGD) —- (iii)
Adding equations (ii) and (iii),
Area (△ABD) + area (△ACD) = 3 area (△BGD) + 3 area (△CGD)
Area (△ABC) = 3 [area (△BGD) + area (△CGD)]
Area (△ABC) = 3 area (△BGC)
1/3 area (△ABC) = area (△BGC) (Hence Proved)
(Q6) The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of it’s altitudes be 6:5:4. Find the lengths of it’s sides.
Solutions:
Given: P = 37 (Perimeter = 37)
Perimeter = x + y + AB
37 = x + y + AB
37 – x – y = AB
Area (△ABC) = 1/2 × x × 6a
= 1/2 × y × 5a
= 1/2 × (37 – x – y) × 4a
= 6x = 5y = (37 – x – y) 4
6x – 5y
x = 5y/6 —— (i)
6x = (37 – x – y) 4
6/4 x = 37 – x – y
3/2 x = 37 – x – y
3x = 2 (37 – x – y)
3x = 74 – 2x – 2y
3x + 2x + 2y = 74
5x + 2y = 74
Put x = 5/6
5 (5/6 y) + 2y = 74
25/6 y + 2y = 74
25y + 12y/6 = 74
37y/6 = 74
y = 74×6/37
y = 2×6
y = 12
∴ y = 12 put in equation (i),
x = 5/6 × y
= 5/6 × 12
= 5 × 2
x = 10
∴ The lengths of sides of a triangle –
BC = x = 10cm
AC = y = 12cm
AB = 37 – x – y = 37 – 10 – 12 = 37 – 22 = 15 cm
(Q7) In parallelogram ABCD, E is a point in AB and DE meets diagonal AC at point F. If DF : FE = 5:3 and area of △ADF is 60cm2 ; find
(i) Area of △ADE
(ii) If AE : EB = 4:5 ; find the areas of △ADB.
(iii) Also, find area of parallelogram ABCD.
Given: DF : FE = 5:3
Area △ADF = 60 cm2
Find : (i) Area of △ADE
(ii) If AE : EB = 4:5, find the area of △ADB
(iii) Also, find area of parallelogram ABCD
Solution:
In △ADE,
Area(△ADF)/area(△ADE) = 5x/8x
60/area(△ADE) = 5/8
60×8/5 = area (△ADE)
12×8 = area (△ADE)
96 cm2 = area (△ADE)
(ii) In △ADB,
Area(△ADE)/area(△ADB) = 4x/9x
96/area (△ADB) = 4/9
96×9/4 = area (△ADB)
24×9 = area (△ADB)
216 cm2 = area (△ADB)
(iii) area (ABCD) = 2×area (△ADB)
= 2 × 216
= 432 cm2
(Q8) In the following figure, BD is parallel to CA, E is mid-point of CA and BD = 1/2 CA. Prove that: area (△ABC) = 2×Area (△DBC)
Solution:
Given: E is mid-point of A and BD = 1/2 CA
To Prove:
Area (△ABC) = 2×area (△DBC)
Proof: area (△BDC) = area (△BEC)
Area (△DBC) = 1/2 area (△ABC)
2 area (△DBC) = area (△ABC) {∵ DE is median} Hence Proved
(Q9) In the following figure, OAB is a triangle and AB = DC
If the area of △CAD = 140 cm2 and the area of △ODC = 172 cm2 ; find
(i) The area of △DBC
(ii) The area of △OAC
(iii) The area of △ODB
Given: AB parallel DC (AB||DC)
Area (△CAD) = 140 cm2
Area (△ODC) = 172 cm2
Find: (i) area (△DBC)
(ii) area (△OAC)
(iii) area (△ODB)
Solution:
Area (△CAD) = area (△DBC)
140 cm2 = area (△BDC) {Triangles are on the same base CD and between same parallel lines)
(ii) Area (△BAC) = area (△OCD) + Area (△CAD)
= 172 + 140
= 312 cm2
(iii) area (△ODB) = area (△OCD) + area (△CBD)
= 172 + 140
= 312 cm2
(Q10) E, F, G and H are the mid-points of the sides of a parallelogram ABCD. Show that area of quadrilateral EFGH is half of the area of parallelogram ABCD.
Solution:
To Prove:
Area (EFGH) = 1/2 area (ABCD)
Proof: In AEGD,
Area (△EHG) = 1/2 area (AEGD) —– (i)
In EBCG,
Area (△EFG) = 1/2 are (EBCG) —– (ii)
Adding equation (i) and (ii),
Area (△EHG) + area (△EFG)
1/2 (area (AEGD) + area (EBCG)]
Area (EFGH) = 1/2 area (ABCD) (Hence Proved)
(Q11) ABCD is a trapezium with AB parallel to DC. A line parallel to AC intersects AB at x and BC at y. Prove that area of △ADX = area of △ACY.
Solution:
To prove:
Area (△ADX) = area (△ACY)
Proof: area (△ADX) = area (△ACX) —— (i)
Also, area (△ACX) = area (△ACY) —- (ii)
From equation (i) and (ii)
Area (△ADX) = area (△ACY) (Hence Proved)
Here is your solution of Selina Concise Class 9 Maths Chapter 16 Area Theorems Exercise 16C
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