Advertisements
Advertisements
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.
Advertisements
उत्तर
Let the height of the larger triangle be h1 and that of the smaller triangle be h2.
The ratio of the areas of two triangles with a common base is equal to the ratio of their corresponding heights.
`"Area(larger Triangle)"/"Area(smaller Triangle)"="h"_1/"h"_2`
`14/9=7/"h"_2`
`14xx "h"_2=9xx7`
`therefore"h"_2=(9xx7)/14=9/2`
`therefore"h"_2=4.5" cm"`
The corresponding height of the smaller triangle is 4.5 cm.
APPEARS IN
संबंधित प्रश्न
In the following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).
In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`
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.
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 figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
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)`
If ΔXYZ ~ ΔPQR then `"XY"/"PQ" = "YZ"/"QR"` = ?
In fig., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?
In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?
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 [______]
`("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`,
`100/110 = (["______"])/"NR"`,
NR = [ ______ ] cm
In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:
- Draw two triangles, give the names of all points, and show heights.
- Write 'Given' and 'To prove' from the figure drawn.
If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.
Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2 xx square xx square`
A(ΔPBC) = `1/2 xx square xx square`
Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
