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In the figure given below, PQR is a triangle with sides PQ=10, PR=17, QR=21. ABCD is a square inscribed in the triangle. I want to find perimeter of square ABCD that is to find the length of side AB. But by using of basic high school geometry concepts, not by trigonometry.

I have drop perpendicular to side QR, and by using heron's formula i found its length 8. but i am confused what to do next. So, please help me.

Any other solutions are expected with above limitation(to use basic high school concepts not by trignometry)

THANKS...............

enter image description here

Community
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curious_mind
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3 Answers3

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Area of Triangle $PQR = 84$ by using semiperimeter formula. From this we get height of Triangle as 8.

Say, side of square $= x$, hence $AB=BC=CD=AD=x$

Now take small triangle $APB$, in this height will be $8-x$. Area of Triangle = $\frac{x\cdot(8-x)}{2}$ -----(1)

Now, Area of Trapezoid $ABRQ = \frac{(21+x)\cdot x}{2}$ -------(2)

As we know, adding (1) & (2), we get area of Triangle $PQR$ i.e 84.

Hence, $$8x-x^2+21x+x^2 = 84\cdot2$$ $$29x = 168$$ $$x = 5.8$$

Hence Perimeter of Square $ABCD= 4x = 4\cdot5.8= 23.2.$

S.D.
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Dhirendra Dhande
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We start as you did. Use Heron's Formula to find the area of the triangle. It is $(12)(7)$.

From that we find, as you did, that the triangle has height $8$.

Drop a perpendicular from the top to the bottom, meeting the bottom at $W$. This divides the big triangle into two Pythagorean triangles with bases $6$ and $15$. (The numbers are "nice." But even if not so nice, we could have found them by using the Pythagorean Theorem.)

Let $x$ be the side length of the square. The point $W$ divides the bottom of the square into two parts. Let $s$ be the length of the left part, and $t$ the length of the right part. Then $s+t=x$.

By similar triangles,

$$\frac{8}{6}=\frac{x}{6-s}=\frac{s+t}{6-s}.$$

By similar triangles, $$\frac{8}{15}=\frac{x}{15-t}=\frac{s+t}{15-t}.$$

We end up with two linear equations in $s$ and $t$. Solve. Then $x=s+t$, so we can find the perimeter of the square.

André Nicolas
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Let $AB=x$. Then the area of trapezoid $QABR=\dfrac{x(21+x)}2$. The area of triangle $ABP=\dfrac{x(8-x)}2$. The sum of these is the area of the whole triangle, which you have already calculated. Solve for x.

EDIT: More simply, note that since the little triangle ABP is similar to the whole triangle, it's base, $AB=\dfrac{21(8-x)}8$. So set this equal to x and solve.

Bill Kleinhans
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