Advertisements
Advertisements
प्रश्न
In fig., PM = 10 cm, A(ΔPQS) = 100 sq. cm, A(ΔQRS) = 110 sq. cm, then NR?
ΔPQS and ΔQRS having seg QS common base.
Areas of two triangles whose base is common are in proportion of their corresponding `square`.
`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,
`100/110 = (square)/(NR)`,
NR = `square` cm
Advertisements
उत्तर
ΔPQS and ΔQRS having seg QS common base.
Areas of two triangles whose base is common are in proportion of their corresponding \[\boxed{\text{heights}}\].
`(A(ΔPQS))/(A(ΔQRS))` = \[\frac{{\boxed{PM}}}{{{NR}}}\],
`100/110` = \[\frac{{\boxed{10}}}{{{NR}}}\],
∴ NR = `(110 xx 10)/100`
∴ NR = \[\boxed{11}\] cm
APPEARS IN
संबंधित प्रश्न
The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.
In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).
The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.
In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:
In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.
- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.
In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.
`"A(∆ ABD)"/"A(∆ ADC)"`
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?
In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?
In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.
In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.
\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\]
A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`
In fig., TP = 10 cm, PS = 6 cm. `(A(ΔRTP))/(A(ΔRPS))` = ?
In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?
In ΔABC, B-D-C and BD = 7, BC = 20, then find the following ratio.
(i) `(A(ΔABD))/(A(ΔADC))`
(ii) `(A(ΔABD))/(A(ΔABC))`
(iii) `(A(ΔADC))/(A(ΔABC))`
