The area of the larger triangle is [tex]\boxed{49{\text{ c}}{{\text{m}}^2}}[/tex] and the area of the smaller triangle is [tex]\boxed{16{\text{ c}}{{\text{m}}^2}}.[/tex]
Further explanation:
Given:
The ratio of the perimeters of two similar triangles is [tex]4:7.[/tex]
Explanation:
If triangles are similar then the ratio of the areas is equal to the square ratio of corresponding sides as well as the square of the perimeter.
Assume that the area of larger triangle as Q.
Assume that the area of smaller triangle as P.
The given ratio of the perimeters of the triangle is [tex]4:7.[/tex]
The ratio of the areas of the larger and smaller triangle can be expressed as follows,
[tex]\begin{aligned}\frac{{{\text{Area of smaller triangle}}}}{{{\text{Area of larger triangle}}}} &= \frac{{{{\left( {{\text{Perimeter of smaller triangle}}} \right)}^2}}}{{{{\left( {{\text{Perimeter of larger triangle}}} \right)}^2}}}\\\frac{{\text{P}}}{{\text{Q}}} &= \frac{{{4^2}}}{{{7^2}}}\\\frac{{\text{P}}}{{\text{Q}}} &= \frac{{16}}{{49}} \\ {\text{49P}} &= {\text{16Q}}\\\end{aligned}[/tex]
The sum of area of larger triangle and the area of smaller triangle is [tex]65{\text{ c}}{{\text{m}}^2}.[/tex]
[tex]\begin{aligned}{\text{P}} + {\text{Q}}&= 65\\{\text{Q}}& = 65 - {\text{P}}\\\end{aligned}[/tex]
Now, substitute [tex]65 - P[/tex] for [tex]Q[/tex] in equation [tex]49P = 16Q.[/tex]
[tex]\begin{aligned}49{\text{P}} &= 16{\text{Q}}\\49{\text{P}} &= 16 \times \left( {65 - {\text{P}}} \right)\\49{\text{P}} &= 16 \times 65 - 16{\text{P}}\\49{\text{P}} + 16{\text{P}} &= 16 \times 65\\65{\text{P}} &= 16 \times 65\\{\text{P}} &= \frac{{16 \times 65}}{{65}}\\{\text{P}} &= 16{\text{ c}}{{\text{m}}^2}\\\end{aligned}[/tex]
The area of larger triangle can be calculated as follows,
[tex]\begin{aligned}{\text{Q}} &= 65 - {\text{P}}\\&= 65 - 16\\&= 49{\text{ c}}{{\text{m}}^2}\\\end{aligned}[/tex]
The area of the larger triangle is [tex]\boxed{49{\text{ c}}{{\text{m}}^2}}[/tex] and the area of the smaller triangle is [tex]\boxed{16{\text{ c}}{{\text{m}}^2}}.[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Triangles
Keywords: Ratio, perimeter, similar, similarity, triangles, proportional, square, area, area of triangle, two similar triangles, sum, sum of areas, 65 cm2, 4:7, corresponding sides.
