Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1 Practice Set 1.1 – Similarity
Balbharati Maharashtra Board Class 10 Math Part – 2 Solution Chapter 1: Similarity. Marathi or English Medium Students of Class 10 get here Similarity full Exercise Solution.
Std |
Maharashtra Class 10 |
Subject |
Math Part 2 Solution |
Chapter |
Similarity |
Practice Set |
1.1 |
Practice Set – 1.1
(1) Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Given, triangle 1 : base = 9, height = 5
∴Area of ∆1 = ½ × b × h
= ½ x 9 x 5
= 45/2
Triangle 2 : base = 10, height = 6
Area of ∆2 = 1/2 × b × h
= 1/2 × 10 × 6
= 30
∴ Area of ∆1 : area of ∆2 = 45/2 × 1/30
= 45/60 = 3/4 = 3:4
(2)In figure 1.13 BC⊥AB, AD⊥AB,
BC = 4, AD = 8, then find A(∆ABC)/A(∆ADB)
Solution:
In the adjacent fig, it is given,
BC⊥AB, AD⊥AB,
BC = 4, AD = 8
AB is the common base for ∆ABC and ∆ABD
A (∆ABC)/A(∆ABD) = BC/AD [∵ bases are equal]
= 4/8
= 1/2
(3) In adjoining figure 1.14 seg PS ⊥seg RQ seg QT ⊥seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.
Solution:
In ∆PQR
With PR as base
A (∆PQR) = 1/2 × b × h
= 1/2 × PR × QT
= 1/2 × 12 × QT
= 6QT
With RQ as base
A (∆PRQ) = 1/2 × RQ × PS
= 1/2 × 6 × 6
= 18
[∵ RQ is the collinear segment of lin RS and PS is the perpendicular from line RS to Vertice P]
∴ 6QT = 18
Or, QT = 18/6 = 3
(4) In adjoining figure, AP⊥BC, AD||BC, then find A (∆ABC) : A (∆BCD).
Solution:
In the adjacent fig it is given that,
AD||BC
AP⊥BC
∆ABC has equal base and equal height – since AP is the height – of ∆ABC and ∆ABC and ∆BPC lies within the same parallel lines with same base.
∴A(∆ABC)/A(∆BDC) = 1
[∵both∆ has equal base and equal height – so they have equal area].
(5) In adjoining figure PQ⊥BC, AD⊥BC then find following ratios.
(i) A(∆PQB)/A(∆PBC)
(ii) A(∆PBC)/A(∆ABC)
(iii) A(∆ABC)/A(∆ADC)
(iv) A(∆ADC)/A(∆PQC)
Solution:
In the adjacent fig, it is given that, PQ⊥BC, AD⊥BC.
∆PQB & ∆PBC has same height – PR
(i) A(∆PQB)/A(∆PBC) = BQ/BC
[∵both height has same height so areas are perpendicular to their base]
(ii) ∆PBC & ∆ABC has same base BC
∴A(∆PBC)/A(∆ABC) = PQ/AD
[∵Both triangle has same base so areas are proportional to their height]
(iii) ∆ABC and ∆ADC has same height.
∴A(∆ABC)/A(∆ADC) = BC/DC [∵ both triangle has same height so their areas are proportional to their base]
(iv) A (∆ADC) = ½ × DC × AD
A (∆PQC) = 1/2 × QC × PQ
∴A(∆ADC)/A(PQC) = (1/2 × DC × AD)/(1/2 × QC × PQ) = DC×AD/QC×PQ
Here is your solution of Maharashtra Board Class 10 Math Part – 2 Chapter 1 Similarity Practice Set 1.1
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