Selina Concise Class 9 Maths Chapter 16 Area Theorems Exercise 16B Solutions
EXERCISE – 16B
(Q1) Show that:
(1) (i) A diagonal divides a parallelogram into two triangles of equal area.
(ii) The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
(iii) The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.
Solution:
(i) Proof:
In △ADC and △ABC
∠DAC = ∠BCA (Alternate interior angles)
AC = AC (Common side)
∠DCA = ∠BAC (Alternate interior angles)
∴ △ADC ≅ △ABC (by angle-side-angle (ASA))
Then, area (△ADC) = area (△ABC)
(ii) Proof:
Area (△ABD) = 1/2 × BD × AE —— (i)
Area (△ADC) = 1/2 × CD × AE ——- (ii)
From equation (i) and (ii),
Area (△ABD)/area (△ADC) = (1/2×BD×AE)/(1/2×CD×AE)
Area (△ABD)/area (△ADC) = m/n (Hence Proved)
(iii) Proof:
Area (△ADB) = 1/2 × AB × DF —– (i)
Area (△ACB) = 1/2 × AB × CE —– (ii)
Equation (i) divided by equation (ii), ((i) ÷ (ii))
Area (△ADB)/area (△ACB) = (1/2×AB×DF)/(1/2×AB×CE)
Area (△ADB)/Area (△ACB) = DF/CE (Hence Proved)
(Q2) In the given figure; AD is median of △ABC and E is only point on median AD. Prove that Area (△ABE) = Area (△ACE)
Solution:
Given: AD is median.
To Prove:
Area (△ABE) = area (△ACE)
Proof: In △ABC,
Area (△ABD) = area (△ACD) —– (i)
In △BEC,
Area (△BED) = area (△CED) —– (ii)
Subtracting equation (i) & (ii),
Area (△ABD) – area (△BED) = area (△ACD) – area (△CED)
Area (△ABE) = area (△ACE) (Hence Proved)
(Q3) In the figure of question 2, if E is the mid-point of median AD, then prove that:
Area (△ABE) = 1/4 area (△ABC).
Solution:
If E is mid-point of median AD.
To Prove:
Area (△ABE) = 1/4 area (△ABC)
In △ABC,
Area (△ABD) = area (△ACD) —– (i)
In △BEC, area (△BED) = area (△CED) —– (ii)
In △ABD,
Area (△ABE) = area (△BED) —– (iii)
From equation (i), (ii) and (iii),
Area (△ABE) = area (△BED) = area (△CBD) = area (△ACE)
= x unit2
Now, area (△ABC) = area (△ABE) + area (△BED) + area (△CED) + area (△ACE)
= x + x + x + x
Area (△ABC) = 4x
1/4 area (△ABC) = x
1/4 area (△ABC) = area (△ABE) (Hence Proved)
(Q4) ABCD is a parallelogram, P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = 1/8 of the area of parallelogram ABCD.
Solution:
Given, ABCD is a parallelogram P and Q are the mid-points of sides AB and AD.
To prove:
Area (△APQ) = 1/8 area (ABCD)
Proof:
Area (ABCD) = area (△ABD) + area (△BCD)
Area (ABCD) = area (△ABD) + area (△ABD)
Area (ABCD) = 2 area (△ABD) —- (i)
Also, area (△BDP) = area (△ACP) —— (ii)
Now, area (ABCD) = 2 [area (△BDP) + area (△ADP)] [From equation (i)]
Area (ABCD) = 4 area (△ADP)
Area (ABCD) = 4 [area (△APQ) + area (△QDP)
Area (ABCD) = 4 [area (△APQ) + area (△APQ)] [area APQ = area QD]
Area (ABCD) = 8 area (△APQ)
1/8 area (ABCD) = area (△APQ) (Hence Proved)
(Q5) The Base BC of △ABC is divided at D so that BD = 1/2 DC.
Prove that area of △ABD = 1/3 rd of the area of △ABC.
Solution:
Given:
To prove:
Area (△ABD) = 1/3 area (△ABC)
Proof: BD = 1/2 DC
BD/DC = 1/2
Let, BD = 1x and CD = 2x
Area (△ABD) = 1/2 × BD × AE —– (i)
Area (△ABC) = 1/2 × BC × AE —— (ii)
Divided equation (i) by (ii), ((i) ÷ (ii))
Area (△ABD)/area (△ABC) = (1/2×BD×AE)/(1/2×BC×AE)
Area(△ABD)/area(△ABC) = BD/BC
Area(△ABD)/area(△ABC) = x/3x
Area (△ABD) = 1/3 area (△ABC) (Hence Proved)
(Q6) In a parallelogram ABCD, Point P lies in DC. Such that DP:PC = 3:2. If area of △DPB = 30 sq. cm, find the parallelogram ABCD.
Solution:
Given:
DP/PC = 3/2
Area (△DPB) = 30 cm2
To find: Area (ABCD) = ?
Find: area (△DBP)/area(△BPC) = 3/2
30/area(△BPC) = 3/2
30×2/3 = area (△BPC)
10×2 = area (△BPC)
20 = area (△BPC)
Now, area (△BDC) = 30+20
= 50 cm2
Therefore, area (ABCD) = 2 × 50
= 100 cm2
(Q7) ABCD is a parallelogram in which BC is produced to E such that CE = BC and AE intersects CD at F. If area (△DFB) = 30 cm2, find the area of parallelogram.
Given:
Area (△DFB) = 30 cm2
Find: area (ABCD) =?
Solution:
In △ADF and △ECF
∠FAD = ∠FEC (Alternate interior angles)
AD = CE (∵ Given)
∠ADF = ∠ECF (Alternate interior angles)
∴ △ADF ≅ △ECF (By Angle-side-Angle (ASA)
DF = FC (By CPCTC)
In △DBC,
Area (△DBF) = area (△FBC) = 30° {because BF is median}
Then, area (△DBC) = 30 + 30
= 60 cm2
Hence, area (ABCD) = 2 × area (△DBC)
= 2 × 60
= 120 cm2
(Q8) The following figure shows a triangle ABC in which P, Q and R are mid-points of sides AB, BC and CA respectively. S is mid-point of PQ.
Prove that: Area (△ABC) = 8×area (△QSB)
Solution:
P, Q and R are mid-points of sides AB, BC and CA respectively S is mid-points of PQ.
To Prove: Area (△ABC) = 8 area (△QSB)
Proof: RQ parallel AP (RQ||AP) (By mid-point theorem)
AR parallel QP (AR||QP)
Then, APQR is a parallelogram.
Area (△APR) = area (△PQR) —– (i)
Similarly, area (△QRG) = area (PQR) —– (ii)
Area (△PQR) = area (△QBP) —- (iii)
From equation (i), (ii) and (iii),
Area (△QRC) = area (△PQR) = area (△QBP) = area (△APR) = x unit
Now, area (△ABC) = area (△APR) + area (△QRC) + area (△PQR) + area (△QBP)
Area (△ABC) = x + x + x + x
Area (△ABC) = 4x
Area (△ABC) = 4 area (△QBP)
Area (△ABC) = 4 [area (△BQS) + area (△BPS)]
= 4 [2 area (△QSB)]
Area (△ABC) = 8 area (△QSB) (Hence Proved)
Here is your solution of Selina Concise Class 9 Maths Chapter 16 Area Theorems Exercise 16B
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